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Let $f(x,y)=xy$ where $x,y\geq 0$. Prove that the function $f$ satisfies the following property: $$f \left(\lambda x + (1- \lambda )x' , \lambda y +(1- \lambda )y' \right) \geq \min \{f(x,y), f(x',y')\}$$ for all $(x,y) \neq (x',y')$ and $ \forall\; \lambda \in (0,1)$

This is a problem from an olympiad book. My try: The left side of the inequality is a quadratic in $\lambda$ i.e. $$ (x-x')(y-y')\lambda^2 + \left\{(x-x')y' + (y-y')x'\right\}\lambda + x'y'$$ I'm unable to find the minimum value of above quadratic in $(0,1)$. Also, how to deal with $\min{\{xy,x'y'\}}$.

Thank you.

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  • $\begingroup$ The equality holds also if $x=x’=0$ or $y=y’=0$. $\endgroup$ Dec 26, 2015 at 18:59
  • $\begingroup$ @AlexRavsky Thanks.. edited the question $\endgroup$ Dec 28, 2015 at 5:16

1 Answer 1

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To simplify, let $m:=\min\{xy,x'y'\}$.

We can write $f \left(\lambda x + (1- \lambda )x' , \lambda y +(1- \lambda )y' \right)$ as:

\begin{eqnarray} f \left(\lambda x + (1- \lambda )x' , \lambda y +(1- \lambda )y' \right) & = & (\lambda x + (1- \lambda )x')(\lambda y +(1- \lambda )y')\\ & = & \lambda^2xy+\lambda(1-\lambda)(xy'+x'y)+(1-\lambda)^2x'y' \end{eqnarray}

Now we prove $xy'+x'y\geq 2m$, looking at all possible cases:

  1. $x\geq x',y\geq y'\Rightarrow xy'\geq x'y', x'y\geq x'y'\Rightarrow xy'+x'y\geq 2x'y'=2m$
  2. $x\leq x',y\leq y'$. Very similar to case 1.
  3. $x\geq x',y\leq y'$. Note $$ 0\geq (x-x')(y-y')=xy-(x'y+xy')+x'y'\\ \Rightarrow x'y+xy'\geq xy+x'y'\geq 2m $$
  4. $x\leq x',y\geq y'$. Same as case 3.

We have $xy'+x'y\geq 2m$ and $xy\geq m, x'y'\geq m$, and since $(x,y)\neq (x',y')$, some of the inequalities will be strict. Hence \begin{eqnarray} f \left(\lambda x + (1- \lambda )x' , \lambda y +(1- \lambda )y' \right) & = & \lambda^2xy+\lambda(1-\lambda)(xy'+x'y)+(1-\lambda)^2x'y'\\ & > & m\lambda^2+2m\lambda(1-\lambda)+m(1-\lambda)^2\\ & = & m(\lambda+(1-\lambda))^2\\ & = & m \end{eqnarray}

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