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if $a,b,c,d$ are positive real numbers that $a+b+c+d=4$,Prove:$$\frac{a}{1+b^2}+\frac{b}{1+c^2}+\frac{c}{1+d^2}+\frac{d}{1+a^2} \ge 2$$ Additional info:I'm looking for solutions and hint that using Cauchy-Schwarz and AM-GM because I have background in them.

Things I have done: Using AM-GM $$LHS=\sum\limits_{cyc}\left(a-\frac{ab^2}{1+b^2}\right)\ge \sum\limits_{cyc}\left(a-\frac{ab^2}{2b}\right)=\sum\limits_{cyc}a-\frac{1}{2}(ab+bc+cd+da)$$

So it lefts to Prove that $$\sum\limits_{cyc}a-\frac{1}{2}(ab+bc+cd+da)\ge 2$$

If I show that $ab+bc+cd+da \le4$ then the upper inequality will be true.But I have problem in showing that.

Daniel Fischer
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user2838619
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  • Am I missing something: "a,b,c,d are positive real numbers and a+b+c+d=4" then a=b=c=d=1 - this is the only way the sum can be true with positive real numbers. It is then trivial to prove the inequality. – Dale M Aug 29 '14 at 06:30
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    @DaleM,I think you made a mistake.there are many Possible solution in Positive Real numbers that $a+b+c+d=4$ is true.for example $a=b=c=\frac{1}{2}$ and $d=\frac{5}{2}$. – user2838619 Aug 29 '14 at 06:34
  • So I did - I read positive integers – Dale M Aug 29 '14 at 06:35

2 Answers2

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You could continue using AM-GM $$ab + bc + cd + da = (a+c)(b+d) \le \frac{(\overline{a+c}+\overline{b+d} )^2}4 = 4$$

Macavity
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For your missing step, just observe $$ (a+b+c+d)^2-4(ab+bc+cd+da)=(a-b+c-d)^2\geq 0\\ \implies ab+bc+cd+da\leq\frac{4^2}{4}=4. $$

Kim Jong Un
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