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I am trying to prove that if $a>1$ $x,y>0$, and there is $b$ such that $\frac{1}{a} + \frac{1}{b} = 1$ then xy $\leq \frac{x^a}{a} + \frac{y^b}{b}$

I have made several attempts, but I get to no answer. Is there any inequality that needs to be shown before?

Fer Stein
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  • Check out Holder's inequality https://en.wikipedia.org/wiki/H%C3%B6lder%27s_inequality – Mostafa Ayaz Aug 24 '19 at 22:33
  • Please try to make the titles of your questions more informative. For example, Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here. – Shaun Aug 24 '19 at 23:38

2 Answers2

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Here is an elementary proof. Fix $y>0$ and let $f(x)= \frac {x^{a}} a+\frac {y^{b}} b-xy$. Then $f'(x) <0$ for $x <y^{1/(a-1)}$ and $f'(x) >0$ for $x >y^{1/(a-1)}$. It follows that $f$ decreases up to $x =y^{1/(a-1)}$ and then increases. So its minimum value is attained when $x =y^{1/(a-1)}$. A simple computation shows that $f(x)=0$ when $x =y^{1/(a-1)}$. Hence $f(x) \geq 0$ for all $x$ which is the given inequality.

Kavi Rama Murthy
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The result you are looking for is Young's inequality for products. Let $a, b >0$. Then, let $t = 1/a$ and $1-t = 1/b$. Then $$x^a / a + y^b / b = tx^a + (1-t)y^b .$$

Taking the natural log of the righthand side of the expression, we get $$\ln(tx^a + (1-t)y^b) \geq t\ln(x^a) +(1-t)\ln(y^b) = \ln(x) + \ln(y) = \ln(xy).$$

Exponentiating both sides of this give the desired relation.

kkc
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