# Using Young's inequality to prove $\frac{|x|^{a_1}|y|^{a_2}}{|x|^{b_1}+|y|^{b_2}} \geq 1$ where $\frac{a_1}{b_1}+\frac{a_2}{b_2}=1$

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?

• The above is wrong - most likely the inequality is the other way - as $x=1, y$ near $0$ shows – Conrad May 18 at 21:45
• @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. – zuotiandenixi May 19 at 1:07

## 1 Answer

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, 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!