show/hide this revision's text 3 added 9 characters in body

The following proof was inspired by Fedor Petrov's and Gjergji's Zaimi's argument, but it is simpler.

By a scaling argument we may assume $a_i\in[1,A]$, $b_i\in[1,B]$. The inequality can be rewritten as $$x^{1/p}y^{1/q} \leq (A^pB^q-1)\sum_{i=1}^n a_ib_i,$$ where $$x:=p(AB^q-B)\sum_{i=1}^na_i^p\qquad\text{and}\qquad y:=q(BA^p-A)\sum_{i=1}^nb_i^q.$$ By Young's inequality $x^{1/p}y^{1/q}\le \frac{x}{p}+\frac{y}{q}$, the above follows from $$\frac{x}{p}+\frac{y}{q}\leq (A^pB^q-1)\sum_{i=1}^n a_ib_i.$$ Therefore it suffices to show, for any $i$, $$(AB^q-B)a_i^p+(BA^p-A)b_i^q\leq (A^pB^q-1)a_ib_i.$$ The difference LHS-RHS is a convex function of $a_i$ and $b_i$, hence we can assume that $a_i\in\{1,A\}$, $b_i\in\{1,B\}$. The inequality is trivial becomes an identity when exactly one of $a_i$ and $b_i$ equals 1, while in the other two cases it is equivalent to $$(1-A^{1-p})(1-B^{1-q})\leq(A-1)(B-1)\leq (A^p-A)(B^q-B).$$ By convexity again, $$1-A^{1-p}\leq(p-1)(A-1)\leq A^p-A,$$ $$1-B^{1-q}\leq(q-1)(B-1)\leq B^q-B,$$ whence the required inequality follows upon noting that $(p-1)(q-1)=1$.
show/hide this revision's text 2 Small change in the endgame.

The following proof was inspired by Fedor Petrov's and Gjergji's Zaimi's argument, but it is simpler.

By a scaling argument we may assume $a_i\in[1,A]$, $b_i\in[1,B]$. The inequality can be rewritten as $$x^{1/p}y^{1/q} \leq (A^pB^q-1)\sum_{i=1}^n a_ib_i,$$ where $$x:=p(AB^q-B)\sum_{i=1}^na_i^p\qquad\text{and}\qquad y:=q(BA^p-A)\sum_{i=1}^nb_i^q.$$ By Young's inequality $x^{1/p}y^{1/q}\le \frac{x}{p}+\frac{y}{q}$, the above follows from $$\frac{x}{p}+\frac{y}{q}\leq (A^pB^q-1)\sum_{i=1}^n a_ib_i.$$ Therefore it suffices to show, for any $i$, $$(AB^q-B)a_i^p+(BA^p-A)b_i^q\leq (A^pB^q-1)a_ib_i.$$ The difference LHS-RHS is a convex function of $a_i$ and $b_i$, hence we can assume that $a_i\in\{1,A\}$, $b_i\in\{1,B\}$. The resulting 4 inequalities are all inequality is trivial , except the when exactly one for of $a_i=b_i=1$ which a_i$ and $b_i$ equals 1, while in the other two cases it is equivalent to $$(A-1)(B-1)\leq $(1-A^{1-p})(1-B^{1-q})\leq(A-1)(B-1)\leq (A^p-A)(B^q-B).$$ By convexity again(or by Bernoulli's inequality), $$(p-1)(A-1)\leq (A^p-A)\qquad\text{and}\qquad (q-1)(B-1)\leq (B^q-B),$$ $1-A^{1-p}\leq(p-1)(A-1)\leq A^p-A,$$ $$1-B^{1-q}\leq(q-1)(B-1)\leq B^q-B,$$ whence the required inequality follows upon noting that $(p-1)(q-1)=1$.
show/hide this revision's text 1

The following proof was inspired by Fedor Petrov's and Gjergji's Zaimi's argument, but it is simpler.

By a scaling argument we may assume $a_i\in[1,A]$, $b_i\in[1,B]$. The inequality can be rewritten as $$x^{1/p}y^{1/q} \leq (A^pB^q-1)\sum_{i=1}^n a_ib_i,$$ where $$x:=p(AB^q-B)\sum_{i=1}^na_i^p\qquad\text{and}\qquad y:=q(BA^p-A)\sum_{i=1}^nb_i^q.$$ By Young's inequality $x^{1/p}y^{1/q}\le \frac{x}{p}+\frac{y}{q}$, the above follows from $$\frac{x}{p}+\frac{y}{q}\leq (A^pB^q-1)\sum_{i=1}^n a_ib_i.$$ Therefore it suffices to show, for any $i$, $$(AB^q-B)a_i^p+(BA^p-A)b_i^q\leq (A^pB^q-1)a_ib_i.$$ The difference LHS-RHS is a convex function of $a_i$ and $b_i$, hence we can assume that $a_i\in\{1,A\}$, $b_i\in\{1,B\}$. The resulting 4 inequalities are all trivial, except the one for $a_i=b_i=1$ which is equivalent to $$(A-1)(B-1)\leq (A^p-A)(B^q-B).$$ By convexity again (or by Bernoulli's inequality), $$(p-1)(A-1)\leq (A^p-A)\qquad\text{and}\qquad (q-1)(B-1)\leq (B^q-B),$$ whence the required inequality follows upon noting that $(p-1)(q-1)=1$.