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Let $u, v, p, q > 0$, and let $\dfrac 1p + \dfrac 1q = 1$. Then

$$uv \le \dfrac {u^p}{p}+ \dfrac {v^q}{q}$$

Show that equality occurs iff $u^p = v^q$.

This is problem $10$ from Ch. 6 in Rudin's PMA (integration chapter). I tried two approaches:

Approach 1:

Solving $\frac 1p + \frac 1q = 1$ for $q$ gives $q = \dfrac {p}{p-1}$. Letting $p=x$ we make the RHS of the inequality into a function: $f(x) = \dfrac {u^x}{x}+\dfrac {v^{\frac {x}{x-1}}}{\frac {x}{x-1}}= \dfrac {u^x}{x}+\dfrac {x-1}{x}v^{\frac {x}{x-1}}$. I plotted $f$ and the results look promising for a first derivative test:

$f'(x) = \dfrac {u^x x \ln u - u^x}{x^2} + \dfrac {1}{x^2}v^{\frac{x}{x-1}} + \dfrac {1}{x^2} + \dfrac {x}{x-1} \cdot \dfrac {1}{x^2} (\ln v) v^{\frac {x}{x-1}}$.

However, I haven't been able to find the solution to $f'(x) = 0$. Yet I know there must be an analytic solution, because the problem tells us that equality occurs iff $u^p = v^q$, so $u^x = v^ {\frac {x}{x-1}}$, and we can solve this equation for $x$ analytically.

Approach 2:

I rewrote the RHS as $\displaystyle \int_0^p u^{t-1} dt + \int_0^q v^{t-1} dt$, but I don't really know what to do from here.

StubbornAtom
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Ovi
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    see here https://math.stackexchange.com/questions/259826/purely-algebraic-proof-of-youngs-inequality – Dr. Sonnhard Graubner May 18 '18 at 16:46
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    This is Young's inequality: https://en.wikipedia.org/wiki/Young%27s_inequality_for_products – Angina Seng May 18 '18 at 16:47
  • For each $u, p, q$ fixed, define $f(v) = \frac{u^p}p + \frac{v^q}{q} -uv$. $f'(v) = v^{q-1} -u$ has unique solution $v = u^{\frac{1}{q-1}}$, $f'(v) < 0$ if $v < u^{\frac{1}{q-1}}$ and $f'(v) >0$ if $v > u^{\frac{1}{q-1}}$. Hence $f(v) \geq f( u^{\frac{1}{q-1}}) =0$. – nguyen0610 May 18 '18 at 16:48

1 Answers1

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The standard way is to prove that $$t \le \frac {t^p}{p} + \frac 1q$$ whenever $t \ge 0$. Define $$\phi(t) = \frac {t^p}{p} + \frac 1q - t.$$ Then $\phi'(t) = t^{p-1} - 1$, so by the first derivative test $\phi$ attains its minimum at $t=1$. Thus $\phi(t) \ge \phi(1)$ for all $t \ge 0$ which rearranges to the stated inequality.

Now evaluate the inequality at $t = uv^{1-q}$, then multiply both sides by $v^q$.

Umberto P.
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