An inequality derived from an olympiad one

Question

This problem is a kind of complement of this.

Let $a,b,x,y,z\gt 0$, prove

$$\frac{x+y+z}{a+b}\ge \frac{xy^3}{ax^3+by^3}+\frac{yz^3}{ay^3+bz^3}+\frac{zx^3}{az^3+bx^3}$$

I have verified that this inequality is valid by computer and a lot of particular $(a, b, x, y, z)$ but I could not yet find a non-sophisticated proof.

The numerator of the last term should be $zx^3$? And the link in your question doesn’t work. — Macavity, Mar 01 '19 at 02:50
Thank you very much. The link is https://math.stackexchange.com/questions/1775572/olympiad-inequality-sum-limits-cyc-fracx48x35y3-geqslant-fracxy/2424065?noredirect=1#comment6439072_2424065 — Piquito, Mar 01 '19 at 03:02

score 1 · Accepted Answer · answered Mar 01 '19 at 03:11

1

It's wrong.

Try $(a,b,x,y,z)=(2,1,3,1,2).$

But it seems that it's true for $0< a\leq b$

answered Mar 01 '19 at 03:11

Michael Rozenberg

194,933

Thanks a lot for your counter example – Piquito Mar 01 '19 at 03:28
@Piquito You are welcome! I proved your inequality for $a=b=1$, but I think there is a hard inequality for $a>b$, which still is true. – Michael Rozenberg Mar 01 '19 at 03:30
@Michel Rozenberg: It is valid for $(a,b)=(8,5)$ and in fact you do have $$A\ge\frac{x+y+z}{13}\ge B$$ where $A$ is the LHS in the link and $B$ is the RHS here. (I have written this problem is derived from the link). – Piquito Mar 01 '19 at 03:39
@Piquito Yes, of course! I see it. – Michael Rozenberg Mar 01 '19 at 03:41

An inequality derived from an olympiad one

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