4
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
  • 3
    $\begingroup$ 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
  • 1
    $\begingroup$ @Gribouillis equal to 1 is not needed $\endgroup$ – N.S.JOHN Dec 26 '16 at 8:06
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ With computer, BW kills it. Michael Rozenberg knew this. $\endgroup$ – River Li Sep 6 at 15:13
0
$\begingroup$

HINT

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 $r + \frac 1 r \ge 2, r \gt 0$.

$\endgroup$
  • 2
    $\begingroup$ Show please, how you use $r+\frac{1}{r}\geq2$. Thank you! $\endgroup$ – Michael Rozenberg Dec 26 '16 at 8:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.