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How to prove inverse of Hölder inequality?

Let $p,q>0,a,b,x,y>0$, and such $$\dfrac{1}{p}+\dfrac{1}{q}=1$$ show that $$\left(a^p+b^p\right)^{\frac{1}{p}}\left(x^q+y^q\right)^{\frac{1}{q}}\le \max{(ax,by)}+\max{(ay,bx)}$$

For this inequality I can't have any idea to do,because this right hand is strange;

Community
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  • $\dfrac{1}{p}+\dfrac{1}{q}=1$! –  Jul 16 '15 at 07:46
  • This inequality is invariant with respect to rescaling of $(a, b)$ and of $(x, y)$. (That is, the transformations $a=\lambda \tilde{a},\ b=\lambda\tilde{b}$ and $x=\mu\tilde{x},\ y=\mu\tilde{y}$, where $\lambda > 0$ and $\mu>0$, leave the inequality unchanged). You can exploit this symmetry. This means that it is enough to prove the inequality with the additional assumptions $a^p+b^p=1$ and $x^q+y^q=1$. – Giuseppe Negro Jul 16 '15 at 08:03
  • @ConĐườngNghệ is $p$ supposed to be $>1$? – Surb Aug 09 '15 at 16:06
  • @Surb $p,q>0$ and $1/p+1/q=1$ imply $p>1$. – David C. Ullrich Aug 10 '15 at 16:35
  • @ConĐườngNghệ, it is related to optimization theory " convexity notion" – zeraoulia rafik Aug 10 '15 at 16:47

2 Answers2

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Wlog $x\ge y$ and $a\ge b$. Divide both sides by $ax$. Saying $s=y/x$ and $t=b/a$, we need to show that $$(1+t^p)^{1/p}(1+s^q)^{1/q} \le\max(1,st)+\max(s,t)=1+\max(s,t)\quad(s,t\in[0,1]).$$But if $A=\max(s,t)$ then $A\le1$, hence $$(1+t^p)^{1/p}(1+s^q)^{1/q} \le(1+A^p)^{1/p}(1+A^q)^{1/q}\le(1+A)^{1/p}(1+A)^{1/q}=1+A.$$

Bonus: We have equality if and only if $x=y$ and $a=b$.

For equality in the first inequality above we need $s=t=A$. Now for equality at the next step we need $A=1$. So we have equality if and only if $s=t=1$.

David C. Ullrich
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Using the convexity of $e^x$ and Jensen's Inequality, we get $$ \begin{align} (1+x)^a(1+y)^{1-a} &=e^{a\log(1+x)+(1-a)\log(1+y)}\\ &\le ae^{\log(1+x)}+(1-a)e^{\log(1+y)}\\ &=1+ax+(1-a)y\tag{1} \end{align} $$

Thus, if we assume $a\ge b$ and $x\ge y$, we have $$ \begin{align} \left(a^p+b^p\right)^{\frac1p}\left(x^q+y^q\right)^{\frac1q} &=ax\left(1+\left(\frac ba\right)^p\right)^{1/p}\left(1+\left(\frac yx\right)^q\right)^{1/q}\tag{2}\\ &\le ax\left(1+\frac1p\left(\frac ba\right)^p+\frac1q\left(\frac yx\right)^q\right)\tag{3}\\ &\le ax\left(1+\frac1p\frac ba+\frac1q\frac yx\right)\tag{4}\\ &=ax+\frac1pbx+\frac1qay\tag{5}\\[6pt] &\le ax+\max(bx,ay)\tag{6}\\[12pt] &=\max(ax,by)+\max(bx,ay)\tag{7} \end{align} $$ Explanation:
$(2)$: factor $a$ out of the first term and $x$ out of the second
$(3)$: apply $(1)$
$(4)$: $\frac ba,\frac yx\le1$ and $p,q\ge1$
$(5)$: distribution
$(6)$: the convex combination of two numbers is less than the greater of the two
$(7)$: remove the assumptions that $a\ge b$ and $x\ge y$

robjohn
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