Prove $(x^a-1)(y^b-1)\geq (1-x^{-b})(1-y^{-a})$ where $x,y\geq 1$ and $a,b \geq 0$. I am trying to see if the inequality always holds. The RHS of course is always $\leq 1$ whereas the LHS can be greater than 1.
Any suggestions greatly appreciated.
 A: Let $f(x)=(x^a-1)(y^b-1)-(1-x^{-b})(1-y^{-a}).$
Thus, $$f'(x)=ax^{a-1}(y^b-1)-bx^{-b-1}(1-y^{-a})=\frac{x^{a+b}a(y^{a+b}-y^a)-b(y^a-1)}{x^{b+1}y^a}\geq$$
$$\geq \frac{a(y^{a+b}-y^a)-b(y^a-1)}{x^{b+1}y^a}.$$
Let $g(y)=a(y^{a+b}-y^a)-b(y^a-1).$
Thus, $$g'(y)=a(a+b)y^{a+b-1}-a^2y^{a-1}-aby^{a-1}=$$
$$=y^{a-1}\left(a(a+b)y^{b}-a^2-ab\right)\geq y^{a-1}(a(a+b)-a^2-ab)=0,$$
which says that $g$ increases.
Thus, $$g(y)\geq g(1)=0,$$ which says that $f$ increases and we obtain:
$$f(x)\geq f(1)=0.$$
A: The two sides are equal if $x = 1$ or $y = 1$ or $a = 0$ or $b = 0.$
Suppose that $x, y > 1$ and $a, b > 0,$ and define the rectangle:
$$
A = [1, x] \times [1, y] \subset \mathbb{R}^2.
$$
Then
\begin{align*}
x^by^a(x^a - 1)(y^b - 1) & = abx^by^a\int_1^xt^{a-1}\,dt\int_1^yu^{b-1}\,du \\
& = abx^by^a\int_At^{a-1}u^{b-1}\,dt\,du \\
& = ab\int_Ax^by^at^{a-1}u^{b-1}\,dt\,du \\
& > ab\int_At^{a+b-1}u^{a+b-1}\,dt\,du \\
& > ab\int_At^{b-1}u^{a-1}\,dt\,du \\
& = ab\int_1^xt^{b-1}\,dt\int_1^yu^{a-1}\,du \\
& = (x^b - 1)(y^a - 1),
\end{align*}
therefore $(x^a - 1)(y^b - 1) > (1 - x^{-b})(1 - y^{-a}).$
