Holder Inequality Generalized? Is it true that $x^{1-1/p}y^{1/p}+z^{1-1/q}w^{1/q}\leq (x+z)^{1-1/(a+b)}(y+w)^{1/(a+b)}$?

Question

I am working on a certain problem and a way to solve it would be to prove an inequality similar to the Holder inequality for sums. The inequality is:

Let $p,q>1$ and $a,b\geq1$ such that $$\dfrac{a}{p}+\dfrac{b}{q}=1.$$ Then, for any positive real numbers $x,y,z$ and $w$ we have $$x^{1-\frac{1}{p}}y^{\frac{1}{p}}+z^{1-\frac{1}{q}}w^{\frac{1}{q}}\leq (x+z)^{1-\frac{1}{a+b}}(y+w)^{\frac{1}{a+b}}.$$

I am not sure whether this inequality actually holds. I have tried many ways to prove it, but I could not conclude any important thing. However I have a little bit of hope that this might be true.

One of the reasons that I think this might be true is because if $p=q=2$ and $a=b=1$ we have a particular case of the Cauchy–Schwarz inequality. Also, if $p=q$, then $a+b=p$ and we are in a particular case of the Holder inequality.

But the situations I am really interested in are the cases where $p\neq q$ and $a,b>1$.

I would love to see your thoughts on this problem, thank you.

Answer 1

No. For $p\ne q$ it is too good to be correct. Suppose $p<q$. Notice that if $p<q$ then $p<a+b<q$. Then take $x=z=w=1$ and $y\to\infty$, so left hand side is of order $y^{1/p}$ while the right hand side is of order $y^{1/(a+b)}$ which is smaller.