How to prove this inverse of Holder inequality? 
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;
 A: 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 $s=t=1$.
A: 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 $x\ge y$
