Let $a$, $b$, $c$ and $d$ be positive numbers. Prove that: $$\left(\frac{a+b}{a+b+c}\right)^2+\left(\frac{b+c}{b+c+d}\right)^2+\left(\frac{c+d}{c+d+a}\right)^2+\left(\frac{d+a}{d+a+b}\right)^2\geq\frac{16}{9}$$ I tried C-S and more, but without success.

I am looking for an human proof, which we can use during competition.

  • $\begingroup$ haha. proof is surely written and analysed by a human. even a robot was programmed by a human first in order for it to solve the problem. $\endgroup$ – DeepSea Dec 26 '16 at 7:39
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    $\begingroup$ Did you try the following substitution: $x =a+b+c, y = b+c+d, z = c+d+a, t = a+b+d$, and solve for $a, b, c,d$ in terms of $x,y,z,t$ to "kill" the denominators ? $\endgroup$ – DeepSea Dec 26 '16 at 7:45
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    $\begingroup$ @Gribouillis equal to 1 is not needed $\endgroup$ – N.S.JOHN Dec 26 '16 at 8:06
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    $\begingroup$ I understand that you are an aficionado of inequalities. May I ask you how/where do you find new ideas almost every week ? $\endgroup$ – Jean Marie Dec 26 '16 at 12:48
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    $\begingroup$ With computer, BW kills it. Michael Rozenberg knew this. $\endgroup$ – River Li Sep 6 at 15:13


Using substitutions $x =a+b+c, y = b+c+d, z = c+d+a, t = a+b+d$ we get $$(\frac{a+b}{a+b+c})^2=\frac 1 9(2 + \frac t x - \frac y x -\frac z x)^2$$ Similar for the other terms. Eliminate the squares then use the fact that $r + \frac 1 r \ge 2, r \gt 0$.

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    $\begingroup$ Show please, how you use $r+\frac{1}{r}\geq2$. Thank you! $\endgroup$ – Michael Rozenberg Dec 26 '16 at 8:38

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