Home » Uncategorized » Inequality 28(Russian Winter Olympiad 2006)

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.

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