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To prove this I have a last inequality wich resists to my reasoning. This is it:

Let $a,b,c,x,y,z$ be positive real numbers where $a,b,c$ are the sides of a triangle, and $x,y,z$ are the sides of a triangle. Then we have: $$(a^2+y^2+z^2)(b^2+x^2+z^2)(c^2+y^2+x^2)\geq\frac{2}{9}(x^2+y^2+z^2)(a^2+b^2+c^2+x^2+y^2+z^2)^2+\frac{1}{9}(x^2+y^2+z^2)^2(a^2+b^2+c^2)$$

The inequality is homogeneous so we can impose the condition: $$a^2+b^2+c^2+x^2+y^2+z^2=1$$

But after that I have no idea to prove this.

Thanks a lot.

1 Answers1

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The inequality is false. Indeed, put $a^2=z^2=\frac 12$ and $b^2=c^2=x^2=y^2=\frac 14$. Then $a,b,c$ and $x,y,z$ are sides of right-angled triangles, $a^2+b^2+c^2=x^2+y^2+z^2=1$, so the inequality becomes $\frac {15}{16}\ge 1$.

Update. If $b^2+x^2=a^2+y^2$ and $a^2+z^2=c^2+x^2$ then $a^2-x^2=b^2-y^2=c^2-z^2$. The inequality is homogeneous so we can impose the condition $x^2+y^2+z^2=1$. Let $a^2+b^2+c^2=A$. Then

$$a^2-x^2=b^2-y^2=c^2-z^2=\frac 13\left(a^2-x^2+b^2-y^2+c^2-z^2\right)= \frac {A-1}3.$$

Then the inequality becomes

$\left(\frac {A-1}3+1\right)^3-\frac29(A+1)^2-\frac19A\ge 0$

$ (A+2)^3-6(A+1)^2-3A\ge 0$

$ (A+2)^3-6(A+1)^2-3A\ge 0$

$A^3-3A+2\ge 0$

$(A+2)(A-1)^2\ge 0$

which is true.

Alex Ravsky
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    Thank you for the answer (+1) , I agree with you the inequality is false in this case . A last question : If we suppose $a\geq b\geq c $ , $x\geq y \geq z$ and $c^2+z^2=b^2+y^2$,$b^2+x^2=a^2+y^2$,and $a^2+z^2=c^2+x^2$ have you a counter-example in this case ?And can we remove one of this two conditions ? Thanks for your interest and the time dedicated to me . –  Oct 18 '17 at 07:31
  • @FatsWallers See the update. Feel free to ask more questions, if your need. – Alex Ravsky Oct 19 '17 at 01:39
  • What do you think about my answer ? https://math.stackexchange.com/questions/1806974/any-similar-lagranges-identity-inequality?noredirect=1&lq=1 –  Oct 19 '17 at 17:27