Vrftsjtryk

The exercise is the following. Let $f:Ulongrightarrowmathbb{R}$ be semiconcave on the open set $U$, that is there exists a constant $Kgeq0$ such that
$$
lambda f(x)+(1-lambda)f(y)leq f(lambda x+(1-lambda)y)+frac{1}{2}Klambda(1-lambda)|x-y|^2
$$
for all $x, yin U$ and $lambdain[0, 1]$. Prove that the above inequality is equivalent to the concavity of the map $xmapsto f(x)-frac{1}{2}K|x|^2$. I cannot prove this equivalence. In particular, by the concavity of $xmapsto f(x)-frac{1}{2}K|x|^2$, I obtain the inequality
$$
lambda f(x)+(1-lambda)f(y)leq f(lambda x+(1-lambda)y)+frac{K}{2}(lambda|x|^2+(1-lambda)|y|^2-|lambda x+(1-lambda)y|^2).
$$
How can I get the desired result?

Thank You

Expanding
$$lvert lambda x + (1 - lambda y)rvert^2 = lambda^2 lvert xrvert^2 + 2lambda(1 - lambda)langle x,yrangle + (1 - lambda)^2 lvert yrvert^2$$
you can write
$$lambdalvert xrvert^2 + (1 - lambda)lvert yrvert^2 - lvert lambda x + (1 - lambda) yrvert^2 = (lambda - lambda^2)lvert xrvert^2 - 2lambda(1 - lambda)langle x,yrangle + [(1 -lambda) - (1 - lambda)^2]lvert yrvert^2$$ or $$lambda(1 - lambda)lvert xrvert^2 - 2lambda(1 - lambda)langle x,yrangle + (1 - lambda)lambda lvert yrvert^2$$ The latter expression is the same as $$lambda(1 - lambda)lvert x - yrvert^2$$
You can do something similar for the proof of the other direction.