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Inequality 23(Iran 1996 TST)


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.


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