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# Inequality 23(Iran 1996 TST)

Problem:

Let $\displaystyle x,y,z$ be non-negative real numbers. Prove that

$\displaystyle \frac{1}{(x+y)^2}+\frac{1}{(y+z)^2}+\frac{1}{(z+x)^2}\geq \frac{9}{4(xy+yz+zx)}$.

1st solution:

The inequality is homogeneous, so, if we normalize it we get that $\displaystyle xy+yz+zx=1$.

Doing some manipulation in the left hand side we acquire:

\displaystyle \begin{aligned}4\sum_{cyc}(x+y)^2(x+z)^2&\geq 9(x+y)^2(y+z)^2(z+x)^2\\&\Longrightarrow 4\sum_{cyc}(x^2+1)^2\geq 9\left(x+y+z-xyz\right)^{2}\end{aligned}.

Let us denote by $\displaystyle s$ the $\displaystyle x+y+z$. Then we have that:

\displaystyle \begin{aligned}4\sum_{cyc}(x^2+1)^2&\geq 9\left(x+y+z-xyz\right)^{2}\\&=4\sum_{cyc}x^4+8\sum_{cyc}x^2+12\geq 9(x+y+z-xyz)^2\end{aligned}.

But $\displaystyle s=x+y+z$, so:

• $\displaystyle \sum_{cyc}x^2=s^2-2$
• $\displaystyle \sum_{cyc}x^4=s^4-4s^2+2+4xyzs$.

Thus the previous inequality can be rewritten as $\displaystyle 4(s^4-2s^2+1+4xyzs)\geq 9(s-xyz)^2$.

Now, from Schur’s inequality we know that

$\displaystyle \sum_{cyc}x^4+xyz\sum_{cyc}x\geq \sum_{cyc}xy(x^2+y^2)\Longrightarrow 6xyzs\geq (4-s^2)(s^2-1)$.

So, we come to the conclusion:

\displaystyle \begin{aligned}4(s^4-2s^2+1+4xyzs)-9(s-xyz)^2&=(s^2-4)(4s^2-1)+34xyz-9x^2y^2z^2\\&\geq (s^2-4)(4s^2-1)+33xyz\geq (s^2-4)(4s^2-1)+\frac{11}{2}(4-s^2)(s^2-1)=\frac{3}{2}(4-s^2)(s^2-3)\geq 0\end{aligned}.

2nd solution:

Doing all the manipulations in the left and in the right hand side we only need to prove that

$\displaystyle 4\sum_{sym}x^5y+\sum_{sym}x^4yz+3\sum_{sym}x^2y^2z^2-3\sum_{sym}x^3y^3-2\sum_{sym}x^3y^2z-\sum_{sym}x^4y^2\geq 0$.

But the last one holds because it is equivalent to:

$\displaystyle 3\left(\sum_{sym}x^5y-\sum_{sym}x^3y^3\right)+\left(\sum_{sym}x^5y-\sum_{sym}x^4y^2\right)+2xyz\sum_{cyc}x(x-y)(x-z)\geq 0$,

whose $\displaystyle 2$ first terms hold from Muirhead’s inequality and the last one from Schur’s inequality.

3nd solution:

Let $\displaystyle a=x+y,b=y+z,c=z+x$. Then the inequality takes the following form:

$\displaystyle \left(-a^2-b^2-c^2+2ab+2bc+2ca\right)\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)\geq 9$.

Doing the manipulations in the left hand side we get that

$\displaystyle \left(\sum_{cyc}\frac{2a}{b}+\frac{2b}{a}-4\right)-\sum_{cyc} \left( \frac{a^2}{c^2}+\frac{b^2}{c^2}-\frac{2ab}{c^2}\right)\geq 0$.

Thus we obtain that

$\displaystyle \sum_{cyc}\frac{2}{ab}(a-b)^2-\frac{1}{c^2}(a-b)^2\geq 0$.

From here we obtain that

$\displaystyle \sum_{cyc}\left(\frac{2}{ab}-\frac{1}{c^2}\right)(b-c)^2\geq 0$.

Let us denote by $\displaystyle S_a$ the $\displaystyle \frac{2}{bc}-\frac{1}{a^2}$ and the $\displaystyle S_b,S_c$ similarly.

Without loss of generality assume that $\displaystyle a\geq b\geq c$. From here we have that

$\displaystyle S_a\geq 0\wedge S_a\geq S_b\geq S_c$.

For the end of our proof we only need to show that $\displaystyle b^2S_b+c^2S_c\geq 0$,

which reduces to

$\displaystyle b^3+c^3\geq abc\Longrightarrow b+c\geq a$, Q.E.D.