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Use Young's Inequality prove that if $a_1$, $a_2$, $b_1$,$b_2$ are all positive and $\frac{a_1}{b_1} + \frac{a_2}{b_2} = 1$ then $$\frac{|x|^{a_1}|y|^{a_2}}{|x|^{b_1}+|y|^{b_2}} {\leq} 1$$ for all $( x,y)\in \mathbb{R}^2\backslash (0,0)$.

I have tried to use Young's inequality. After several attempts to manipulate Young's inequality, I still cannot find a place to start the proof.

Can you help me, please?

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    $\begingroup$ The above is wrong - most likely the inequality is the other way - as $x=1, y$ near $0$ shows $\endgroup$ – Conrad May 18 at 21:45
  • $\begingroup$ @Conrad Thank you! It was an mistake by me. But I still have no clue after manipulating the inequality even if it is the other way. $\endgroup$ – zuotiandenixi May 19 at 1:07
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Let $a=|x|^{a_1},b=|y|^{a_2}, p=\frac{b_1}{a_1}, q=\frac{b_2}{a_2}, p,q>1, \frac{1}{p}+\frac{1}{q}=1, |x|^{b_1}+|y|^{b_2}=a^p+b^q$

Young inequality is $ab \le \frac{1}{p}a^p+\frac{1}{q}b^q$, but $p,q>1$ imply

$\frac{1}{p}a^p+\frac{1}{q}b^q<a^p+b^q$, so putting things together we get

$\frac{|x|^{a_1}|y|^{a_2}}{|x|^{b_1}+|y|^{b_2}}=\frac{ab}{a^p+b^q}<1$ so we are done!

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