\section{On a Class of Three-Variable Inequalities}%20

\markboth{Articles}{On a Class of Three-Variable Inequalities}

\vspace{4.2cm}

\subsection{Theorem}

Let $a,b,c$ be real numbers such that $a+b+c=1$.

By the AM-GM inequality,

$$ab+bc+ca\le \ds\f{1}{3},$$

therefore setting $ab+bc+ca=\ds\f{1-a^2}{3}$ $(q\ge 0)$,

we will find the maximum and minimum values of $abc$ in terms of $q$.

If $q=0$, then $a=b=c=\ds\f{1}{3}$,

therefore $abc=\ds\f{1}{27}$.

If $q\ne 0$, then

$(a-b)^2+(b-c)^2+(c-a)^2>0$.

Consider the function

$f(x)=(x-a)(x-b)(x-c)=x^3-x^2+\ds\f{1-q^2}{3}x-abc$.

We have

$$f'(x)=3x^2-2x+\ds\f{1-q^2}{3}$$

with zeros $x_1=\ds\f{1+q}{3}$ and $x_2=\ds\f{1-q}{3}$.

We can see that

$f'(x)<0$ for $x_2<x<x_1$ and $f'(x)>0$ for $x<x_2$ or $x>x_1$.

Furthermore, $f(x)$ has three zeros: $a,b,c$.

Then

$$f\left(\ds\f{1-q}{3}\right)=\ds\f{(1-q)^2(1+2q)}{27}-abc\ge 0$$

and

$$f\left(\ds\f{1+q}{3}\right)=\ds\f{(1+q)^2(1-2q)}{27}-abc\le 0.$$

Hence

$$\ds\f{(1+q)^2(1-2q)}{27}\le abc\le \ds\f{(1-q)^2(1+2q)}{27}$$

and we obtain the following theorem.

{\bf Theorem 1.}

{\it If $a,b,c,$ are arbitrary real numbers such that $a+b+c=1$,

and if $ab+bc+ca=\ds\f{1-q^2}{3}$ $(q\ge 0)$, then holds

$$\ds\f{(1+q)^2(1-2q)}{27}\le abc\le \ds\f{(1-q)^2(1+2q)}{27}.$$

}

Or, more generally,

{\bf Theorem 2.}

{\it Let $a,b,c$ be real numbers such that $a+b+c = p$.

If $ab+bc+ca =\ds\f{p^2-q^2}{3}$ $(q\ge 0)$ and $r=abc$, then

$$\ds\f{(p+q)^2(p-2q)}{27}\le r\le \ds\f{(p-q)^2(p+2q)}{27}.$$

This is a powerful tool because the equality holds if and only if

$$(a-b)(b-c)(c-a)=0.$$

Here are some identities which we can use with this theorem

$$a^2+b^2+c^2=\ds\f{p^2+2q^2}{3}$$

$$a^3+b^3+c^3=pq^2+3r$$

$$ab(a+b)+bc(b+c)+ca(c+a)=\ds\f{p(p^2-q^2)}{3}-3r$$

$$(a+b)(b+c)(c+a)=\ov{p(p^2-q^2)}{3}-r$$

$$a^2b^2+b^2c^2+c^2a^2=\ds\f{(p^2-q^2)^2}{9}-2pr$$

$$ab(a^2+b^2)+bc(b^2+c^2)+ca(c^2+a^2)

=\ds\f{(p^2+2q^2)(p^2-q^2)}{9}-pr$$

$$a^4+b^4+c^4=\ds\f{-p^4+8p^2q^2+2q^4}{9}+4pr.$$

}

 

{\bf Remark 1.}

There is also a geometric proof of this result.

By Viete's formulas we have

$$(x-a)(x-b)(x-c)=x^2-px^2+\ds\f{1}{3}(p^2-q^2)x-r.$$

As we increase $r$ we lower the graph of

$y=x^3-px^2+\ds\f{1}{3}(p^2-q^2)x-r$.

The largest and smallest values of $r$ for which this polynomial will have three real

roots will be at the points where it has double roots, with the maximum of $r$

corresponding to the case where the smaller root is the double root.

In this case we find the double root at $(p\mp q)/3$ and the single root at $(p\pm 2q)/3$

with the upper sign corresponding to the maximum of $r$ and the lower sign the minimum.

{\bf Remark 2.}

This result is sharp in the following sense.

Suppose we are given real numbers $p,q,r$ with $q\ge 0$ satisfying this inequality.

Then there exist real numbers $a,b,c$ such that $a + b + c = p$,

$ab+bc+ca=\ds\f{p^2-q^2}{3}$, and $abc=r$.

{\bf Remark 3.}

If $a,b,c\ge 0$ then we get the additional restrictions that $r\ge 0$

and $p\ge q$.

This remark is used in the last application so it might be worth making

even if you ignore the others.

Alternately, one can state it with all three

inequalities strict.

For a monic cubic polynomial $p(x)=(x-a)(x-b)(x-c)$

the discriminant of the polynomial $p$ is

$D=(a-b)^2(b-c)^2(c-a)^2$.

This is a symmetric polynomial in the roots of $p$, therefore it can be written as a

polynomial in the coefficients of $p$.

In the special case above where

$p(x)=x^3-px^2+\ds\f{1}{3}(p^2-q^2)x-r$

one can compute that

$$D=27\left(r-\ds\f{(p+q)^2(p-2q)}{3}\right)\left(\ds\f{(p-q)^2(p+2q)}{3}-r\right).$$

With this formula Theorem 2 and its converse combine to give the well known

fact that the discriminant is nonnegative if and only if p has only real roots.

\subsection{Applications}

{\bf Problem 1.}

{\it Let $a,b,c$ be positive real numbers such that $a+b+c = 1$.

Prove that

$$\ds\f{1}{a}+\ds\f{1}{b}+\ds\f{1}{c}+48(ab+bc+ca)\ge 25.$$

}

{\bf Solution.}

We can easily check that $q\in [0,1]$, and by using Theorem 2,

\begin{align*}

{\rm LHS}

& =\ds\f{1-q^2}{3r}+16(1-q^2)\ge \ds\f{9(1+q)}{(1-q)(1+2q)}+16(1-q^2)\\

& =\ds\f{2q^2(4q-1)^2}{(1-q)(1+2q)}+25\ge 25.

\end{align*}

The inequality is proved.

Equality holds if and only if $a = b = c = \ds\f{1}{3}$

or $a=\ds\f{1}{2}$, $b=c=\ds\f{1}{4}$

and their permutations.

{\bf Problem 2.}

(Vietnam 2002).

{\it Let $a,b,c$ be real numbers such that $a^2+b^2+c^2=9$.

Prove that

$$2(a+b+c)-abc\le 10.$$

}

{\bf Solution.}

The condition can be written as

$p^2+2q^2=27$.

Using our theorem, we have

$${\rm LHS}=2p-r\le 2p-\ds\f{(p+q)^2(p-2q)}{27}=\ds\f{p(5q^2+27)+2q^3}{27}.$$

We need to prove that

$$p(5q^2+27)\le 270-2q^3.$$

This follows from

$$(270-2q^3)^2\ge p^2(5q^2+27)^2,$$

or, equivalently,

$$27(q-3)^2(2q^4+12q^3+49q^2+146q+219)\ge 0.$$

The inequality is proved.

Equality holds if and only if $a = b = 2$, $c = -1$ and

their permutations.

{\bf Problem 3.}

(Vo Quoc Ba Can).

{\it For all positive real numbers $a,b,c$,

$$\ds\f{a+b}{c}+\ds\f{b+c}{a}+\ds\f{c+a}{b}+11\sqrt {\ds\f{ab+bc+ca}{a^2+b^2+c^2}}\ge 17.$$

}

{\bf Solution.}

Because the inequality is homogeneous, without loss of generality,

assume that $p = 1$.

Then $q\in [0,1]$ and the inequality can be written as

$$\ds\f{1-q^2}{3r}+11\sqrt {\ds\f{1-q^2}{1+2q^2}}\ge 20.$$

By our theorem, it suffices to prove

$$11\sqrt {\ds\f{1-q^2}{1+2q^2}}\ge 20-\ds\f{9(1+q)}{(1-q)(1+2q)}

=\ds\f{-40q^2+11q+11}{(1-q)(1+2q)}.$$

If $-40q^2+11q+11\le 0$,

or

$q\ge \ds\f{11+3\sqrt {209}}{80}$,

it is trivial.

If

$q\le \ds\f{11+3\sqrt {209}}{80}<\ds\f{2}{3}$,

$$\ds\f{121(1-q^2)}{1+2q^2}-\ds\f{(-40q^2+11q+11)^2}{(1-q)^2(1+2q)^2}

=\ds\f{3q^2(11-110q+255q^2+748q^3-1228q^4)}{(1+2q^2)(1-q)^2(1+2q)^2}.$$

Note that the only positive real root of

$11-110q+255q^2+748q^3-1228q^4$

is at $0.748037\ldots >\ds\f{11+3\sqrt {209}}{80}$

and therefore the numerator is positive for $q\le \ds\f{11+3\sqrt {209}}{80}$.

The inequality is proved.

Equality occurs if and only if $a = b = c$.

{\bf Problem 4.}

(Vietnam TST 1996).

{\it Prove that for any $a,b,c\in \mathbb{R}$,

$$(a+b)^4+(b+c)^4+(c+a)^4\ge \ds\f{4}{7}(a^4+b^4+c^4).$$

}

{\bf Solution.}

If $p = 0$ the inequality is trivial, so we will consider the case $p\ne 0$.

Without loss of generality, assume that $p = 1$.

The inequality becomes

$$q^4+4q^2+10-108r\ge 0$$

and by our theorem

\begin{align*}

q^4+4q^2+10-108r

& \ge q^4+4q^2+10-4(1-q)^2(1+2q)\\

& =q^2(q-4)^2+6\ge 0.

\end{align*}

The inequality is proved.

Equality holds only for $a = b = c = 0$.

{\bf Problem 5.}

(Pham Huu Duc, MR1/2007).

{\it Prove that for any positive real numbers $a,b,c$,

$$\sqrt {\ds\f{b+c}{a}}+\sqrt {\ds\f{c+a}{b}}+\sqrt {\ds\f{a+b}{c}}

\ge \sqrt {6\cdot \ds\f{a+b+c}{\sqrt [3]{abc}}}.$$

}

{\bf Solution.}

By H\"older's inequality,

$$\left(\sum_{cyc}\sqrt {\ds\f{b+c}{a}}\right)^2

\left(\sum_{cyc}\ds\f{1}{a^2(b+c)}\right)\ge \left(\sum_{cyc}\ds\f{1}{a}\right)^3.$$

It suffices to prove that

$$\left(\sum_{cyc}\ds\f{1}{a}\right)^3\ge \ds\f{6(a+b+c)}{\sqrt [3]{abc}}

\sum_{cyc}\ds\f{1}{a^2(b+c)}.$$

Let $x=\ds\f{1}{a}$, $y=\ds\f{1}{b}$, $z=\ds\f{1}{c}$,

then the inequality becomes

$$(x+y+z)^3\ge 6\sqrt [3]{xyz}(xy+yz+zx)\sum_{cyc}\ds\f{x}{y+z},$$

or

$$(x+y+z)^3\ge \ds\f{6\sqrt [3]{xyz}(xy+yz+zx)}{(x+y)(y+z)(z+x)}

((x+y+z)^3-2(x+y+z)(xy+yz+zx)+3xyz).$$

By the AM-GM inequality,

$$(x+y)(y+z)(z+x)=(x+y+z)(xy+yz+zx)-xyz

\ge \ds\f{8}{9}(x+y+z)(xy+yz+zx).$$

It remains to prove that

$$4(x+y+z)^4\ge 27\sqrt [3]{xyz}((x+y+z)^3-2(x+y+z)(xy+yz+zx)+3xyz).$$

Setting $p=x+y+z$, $xy+yz+zx=\ds\f{p^2-q^2}{3}$ $(p\ge q\ge 0)$,

the inequality becomes

$$4p^4\ge 9\sqrt [3]{xyz}(p^3+2pq^2+9xyz).$$

Applying our theorem, it suffices to prove that

$$4p^4\ge 9\sqrt [3]{\ds\f{(p-q)^2(p+2q)}{27}}

\left(p^3+2pq^2+\ds\f{(p-q)^2(p+2q)}{3}\right)$$

$$4p^4\ge \sqrt [3]{(p-q)^2(p+2q)}(3p^3+6pq^2+(p-q)^2(p+2q)).$$

Setting $u=\sqrt [3]{\ds\f{p-q}{p+2q}}\le 1$,

the inequality is equivalent to

$$4(2u^3+1)^4\ge 27u^2(4u^9+5u^6+2u^3+1),$$

or

$$f(u)=\ds\f{(2u^3+1)^4}{u^2(4u^9+5u^6+2u^3+1)}\ge \ds\f{27}{4}.$$

We have

$$f'(u)=\ds\f{2(2u^3+1)^3(u^3-1)(2u^3-1)(2u^6+2u^3-1)}{u^3(u^3+1)^2(4u^6+u^3+1)^2}$$

and

$$f'(u)=0\Lr u=\sqrt [3]{\ds\f{\sqrt 3-1}{2}},\ds\f{1}{\sqrt [3]{2}},1.$$

Now, we can easily verify that

$$f(u)\ge \min\left\{f\left(\sqrt [3]{\ds\f{\sqrt 3-1}{2}}\right),f(1)\right\}=\ds\f{27}{4},$$

which is true.

The inequality is proved.

Equality holds if and only if $a = b = c$.

{\bf Problem 6.}

(Darij Grinberg).

{\it If $a,b,c\ge 0$, then

$$a^2+b^2+c^2+2abc+1\ge 2(ab+bc+ca).$$

}

{\bf Solution.}

Write the inequality as

$$6r+3+4q^2-p^2\ge 0.$$

If $2q\ge p$, it is trivial.

If $p\ge 2q$, using the theorem, it suffices to prove that

$$\ds\f{2(p-2q)(p+q)^2}{9}+3+4q^2-p^2\ge 0,$$

or

$$(p-3)^2(2p+3)\ge 2q^2(2q+3p-18).$$

If $2p\le 9$, we have $2q+3p\le 4p\le 18$,

therefore the inequality is true.

If $2p\ge 9$, we have

$$2q^2(2q+3p-18)\le 4q^2(2p-9)\le p^2(2p-9)$$

$$=(p-3)^2(2p+3)-27<(p-3)^2(2p+3).$$

The inequality is proved.

Equality holds if and only if $a = b = c = 1$.

{\bf Problem 7.}

(Schur's inequality).

{\it For all nonnegative real numbers $a,b,c$,

$$a^3+b^3+c^3+3abc\ge ab(a+b)+bc(b+c)+ca(c+a).$$

}

{\bf Solution.}

Because the inequality is homogeneous, we can assume that

$a+b+c=1$.

Then $q\in [0,1]$ and the inequality is equivalent to

$$27r+4q^2-1\ge 0.$$

If $q\ge \ds\f{1}{2}$, it is trivial.

If $q\le \ds\f{1}{2}$,

by the theorem we need to prove that

$$(1+q)^2(1-2q)+4q^2-1\ge 0,$$

or

$$q^2(1-2q)\ge 0,$$

which is true.

Equality holds if and only if $a = b = c$ or $a = b$, $c = 0$ and their permutations.

{\bf Problem 8.}

(Pham Huu Duc).

{\it For all positive real numbers $a,b,c$,

$$\ds\f{1}{a^2+bc}+\ds\f{1}{b^2+ca}+\ds\f{1}{c^2+ab}

\le \ds\f{(a+b+c)^2}{3(ab+bc+ca)}

\left(\ds\f{1}{a^2+b^2}+\ds\f{1}{b^2+c^2}+\ds\f{1}{c^2+a^2}\right).$$

}

{\bf Solution.}

Because the inequality is homogeneous, we may assume that $p=1$.

Then $q\in [0,1]$,

and by the AM-GM and Schur's inequalities, we have

$$\ds\f{(1-q^2)^2}{9}\ge 3r\ge \max\left\{0,\ds\f{1-4q^2}{9}\right\}.$$

After expanding, we can rewrite the given inequality as

$$f(r)=-486(9-q^2)r^3+27(q^6+64q^4-35q^2+24)r^2$$

$$+9(4q^2-1)(11q^4-4q^2+2)r+q^2(1-q^2)^3(2q^4+8q^2-1)\ge 0.$$

We have

$$f'(r)=9-162(9-q^2)r^2+6(q^6+64q^4-35q^2+24)r

+(4q^2-1)(11q^4-4q^2+2)$$

$$f''(r)=54(-54(9-q^2)r+q^6+64q^4-35q^2+24)$$

$$\ge 54(-2(1-q^2)^2(9-q^2)+q^6+64q^4-35q^2+24)

=162(q^6+14q^4+q^2+2)>0.$$

Hence $f'(r)$ is an increasing function.

Now, if $1\le 2q$, then

$$f'(r)\ge f'(0)=9(4q^2-1)(11q^4-4q^2+2)\ge 0.$$

If $1\ge 2q$, then

$$f'(r)\ge f'\left(\ds\f{1-4q^2}{27}\right)=(1-4q^2)(q^2+2)(2q^4+17q^2+6)\ge 0.$$

In each case, $f(r)$ is an increasing function.

If $1\le 2q$, then

$$f(r)\ge f(0)=q^2(1-q^2)^3(2q^4+8q^2-1)\ge 0,$$

and we are done.

If $1\ge 2q$, by our theorem,

$$f(r)\ge f\left(\ds\f{(1+q)^2(1-2q)}{27}\right)$$

$$=\ds\f{1}{81}q^2(2-q)(q+1)^2(6q^3+4q^2-7q+4)(5q^2-2q+2)^2\ge 0.$$

The proof is complete.

Equality holds if and only if $a = b = c$.

{\bf Problem 9.}

(Nguyen Anh Tuan).

{\it Let $x,y,z$ be positive real numbers such that

$xy + yz + zx + xyz = 4$.

Prove that

$$\ds\f{x+y+z}{xy+yz+zx}\le 1+\ds\f{1}{48}((x-y)^2+(y-z)^2+(z-x)^2).$$

}

{\bf Solution.}

Since $x,y,z>0$ and $xy+yz+zx+xyz=4$, there are $a,b,c>0$ such that

$x=\ds\f{2a}{b+c}$, $y=\ds\f{2b}{c+a}$, $z=\ds\f{2c}{a+b}$.

The inequality becomes

$$P(a,b,c)=\ds\f{(a+b+c)^2\ds\sum_{cyc}(a^2-b^2)^2}{(a+b)^2(b+c)^2(c+a)^2}

-\ds\f{6\ds\sum_{cyc}a(a+b)(a+c)}{\ds\sum_{cyc}ab(a+b)}+12\ge 0.$$

Because the inequality is homogeneous we can assume that $p = 1$.

Then

$q\in [0,1]$,

and after some computations, we can write the inequality as

$$f(r)=729r^3+27(22q^2-1)r^2+27(6q^4-4q^2+1)r+(q^2-1)(13q^4-5q^2+1)\le 0.$$

We have

$$f'(r)=27(r(81r+44q^2-2)+6q^4-4q^2+1).$$

By Schur's inequality,

$$81r+44q^2-2\ge 3(1-4q^2)+44q^2-2=1+32q^2>0.$$

Hence $f'(r)\ge 0$, and $f(r)$ is an increasing function.

Then by our theorem,

$$f(r)\!\le\! f\left(\ds\f{(1-q)^2(1+2q)}{27}\right)

=\ds\f{2}{27}q^2(q-1)(q+2)^2(4q^4+14q^3+15q^2-7q+1)\!\le 0.$$

The inequality is proved.

Equality holds if and only if $x = y = z$.

{\bf Problem 10.}

(Nguyen Anh Tuan).

{\it For all nonnegative real numbers $a,b,c$,

$$\sqrt {(a^2-ab+b^2)(b^2-bc+c^2)}

+\sqrt {(b^2-bc+c^2)(c^2-ca+a^2)}$$

$$+\sqrt {(c^2-ca+a^2)(a^2-ab+b^2)}

\ge a^2+b^2+c^2.$$

}

{\bf Solution.}

After squaring both sides, we can rewrite the inequality as

$$2\sqrt {\ds\prod_{cyc}(a^2-ab+b^2)}\left(\sum_{cyc}\sqrt {a^2-ab+b^2}\right)

\ge \left(\sum_{cyc}ab\right)\left(\sum_{cyc}a^2\right)-\sum_{cyc}a^2b^2.$$

By the AM-GM inequality,

$$\sqrt {a^2-ab+b^2}\ge \ds\f{1}{2}(a+b),\

\sqrt {b^2-bc+c^2}\ge \ds\f{1}{2}(b+c),\

\sqrt {c^2-ca+a^2}\ge \ds\f{1}{2}(c+a).$$

It suffices to prove that

$$2\sqrt {\ds\prod_{cyc}(a^2-ab+b^2)}\left(\sum_{cyc}a\right)

\ge \left(\sum_{cyc}ab\right)\left(\sum_{cyc}a^2\right)-\sum_{cyc}a^2b^2.$$

Because this inequality is homogeneous, we can assume $p = 1$.

Then $q\in [0,1]$

and the inequality is equivalent to

$$2\sqrt {-72r^2+3(1-10q^2)r+q^2(1-q^2)^2}\ge 6r+q^2(1-q^2),$$

or

$$f(r)=324r^2-12r(q^4-11q^2+1)-q^2(4-q^2)(1-q^2)^2\le 0.$$

It is not difficult to verify that $f(r)$ is a convex function, then by our theorem,

$$f(r)\le \max\left\{f(0),f\left(\ds\f{(1-q)^2(1+2q)}{27}\right)\right\}.$$

Furthermore,

$$f(0)=-q^2(4-q^2)(1-q^2)^2\le 0$$

$$f\left(\ds\f{(1-q)^2(1+2q)}{27}\right)

=\ds\f{1}{9}q^2(q-1)^3(q+2)(9q^2+q+2)\le 0.$$

Our proof is complete.

Equality holds if and only if $a = b = c$ or $a=t\ge 0$,

$b = c = 0$, and their permutations.

\bigskip

\hfill

{\Large Vo Quoc Ba Can, Vietnam}

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