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# Inequality 28(Russian Winter Olympiad 2006)

Problem:

Let $\displaystyle a,b,c,d$ be positive real numbers such that $\displaystyle a+b+c+d=4$. Prove that

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

Solution:

Let us make the Cauchy reverse technique and apply AM-GM inequality. Then we have that:

• $\displaystyle \frac{a}{1+b^2}=a-\frac{ab^2}{1+b^2}\leq a-\frac{ab}{2}$,
• $\displaystyle \frac{b}{1+c^2}=b-\frac{bc^2}{1+b^2}\leq b-\frac{bc}{2}$,
• $\displaystyle \frac{c}{1+d^2}=c-\frac{cd^2}{1+d^2}\leq c-\frac{cd}{2}$,
• $\displaystyle \frac{d}{1+a^2}=d-\frac{da^2}{1+a^2}\leq d-\frac{da}{2}$.

So, it is enough to prove that

$\displaystyle \sum_{cyc}a-\frac{1}{2}\sum_{cyc}ab\geq 2\Longrightarrow 4\geq ab+bc+cd+da$.

Multiplying by $\displaystyle 4$ we get that $\displaystyle (a+b+c+d)^2\geq 4(ab+bc+cd+da)$, which reduces to the obvious inequality

$\displaystyle (a-b+c-d)^2\geq 0$, Q.E.D.