Define $f:[0,\infty)^2\to\mathbb{R}$ as $f(x,y)=(1+x)^t(1+y)^{1-t}-x^ty^{1-t}$, where $t\in [0,1]$.
Is it true that $f(x,y)\geq 1$ for all $x,y\geq 0$? I'm almost sure that this is the case. A simple simulation study made me more confident of the inequality. I tried to prove it by differentiation. Since $$ \dfrac{\partial f}{\partial x} = t\left(\dfrac{1+y}{1+x}\right)^{1-t}-t\left(\dfrac{y}{x}\right)^{1-t} $$ and $$ \dfrac{\partial f}{\partial y} = (1-t)\left(\dfrac{1+x}{1+y}\right)^{t}-(1-t)\left(\dfrac{x}{y}\right)^{t} $$ So any saddle point (x,y) is such that $x=y$ and $f(x,x)=1$. But, is this a minimum?
Is there another way to prove it? Of course, if it is true...
Hint: This follows from Hölder's inequality