Prove that $f \left(\lambda x + (1- \lambda )x' , \lambda y +(1- \lambda )y' \right) > \min \{f(x,y), f(x',y')\}$ 
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.
 A: 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:


*

*$x\geq x',y\geq y'\Rightarrow xy'\geq x'y', x'y\geq x'y'\Rightarrow xy'+x'y\geq 2x'y'=2m$

*$x\leq x',y\leq y'$. Very similar to case 1.

*$x\geq x',y\leq y'$. Note $$
0\geq (x-x')(y-y')=xy-(x'y+xy')+x'y'\\
\Rightarrow x'y+xy'\geq xy+x'y'\geq 2m
$$ 

*$x\leq x',y\geq y'$. 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}
