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# Inequality 8(Laurentiu Panaitopol-JBMO 2003)

Problem:

If $\displaystyle a,b,c>-1$ prove that

$\displaystyle \frac{1+a^2}{1+b+c^2}+\frac{1+b^2}{1+c+a^2}+\frac{1+c^2}{1+a+b^2}\geq 2$.

Solution:

From the AM-GM inequality we get that

$\displaystyle \frac{1+a^2}{1+b+c^2}+\frac{1+b^2}{1+c+a^2}+\frac{1+c^2}{1+a+b^2}\geq \frac{1+a^2}{1+\frac{b^2+1}{2}+c^2}+\frac{1+b^2}{1+\frac{c^2+1}{2}+a^2}+\frac{1+c^2}{1+\frac{a^2+1}{2}+b^2}$.

Let $\displaystyle x=1+a^2,y=1+b^2,z=1+c^2$.

Then the inequality transforms into

$\displaystyle \frac{x}{\frac{y}{2}+z}+\frac{y}{\frac{z}{2}+x}+\frac{z}{\frac{x}{2}+y}\geq 2$.

Multiplying the above fractions with $\displaystyle x,y,z$ respectively we get that

$\displaystyle \frac{x^2}{\frac{xy}{2}+xz}+\frac{y^2}{\frac{yz}{2}+yx}+\frac{z^2}{\frac{zx}{2}+zy}\geq 2$.

Applying Cauchy-Schwarz inequality we have that

\displaystyle \begin{aligned}\frac{x^2}{\frac{xy}{2}+xz}+\frac{y^2}{\frac{yz}{2}+yx}+\frac{z^2}{\frac{zx}{2}+zy}&\geq \frac{(x+y+z)^2}{xy+yz+zx+\frac{xy+yz+zx}{2}}\geq 2\\&\Longrightarrow (x+y+z)^2\geq 3(xy+yz+zx)\end{aligned}

,which is obviously true. Thus our inequality is proved, Q.E.D.