How to prove convexity + quasilinear preferences imply concave utility? Let $\succsim$ be a strictly convex and quasilinear preference relation. It's defined over, say, $\mathbb{R}^2_{+}$ and is quasilinear on good 1.

So, $U(x_{1},x_{2}) = x_{1} + f(x_{2})$. How to prove that $f$ is a strictly concave function?

I'm solving problem 15.B.8 from MWG and I can't even understand what the solutions manual did! 

Thanks a lot in advance for any hints and ideas!
Consider any $x_2'$ and $x_2''$ in $\mathbb{R}_+$. Without loss of generality, let $x_2'' &gt; x_2'$. We can choose $x_1'=f(x_2'')-f(x_2') &gt; 0$ so that $U(0,x_2'')=U(x_1',x_2')$. Let $\lambda(x_1',x_2')+(1-\lambda)(0,x_2'')$ be a convex combination of $(x_1',x_2')$, and $(0,x_2'')$. Since $\succsim$ is strictly convex and $U(0,x_2'')=U(x_1',x_2')$,
        \begin{eqnarray*} 
 &amp;&amp; U(\lambda(x_1',x_2')+(1-\lambda)(0,x_2''))  &gt;  U(x_1',x_2') \\
 &amp;\Rightarrow &amp; U(\lambda x_1'+(1-\lambda)0,\lambda x_2'+(1-\lambda)x_2'')  &gt;  \lambda U(x_1',x_2')+(1-\lambda)U(0,x_2'')\ldots(\because U(0,x_2'')=U(x_1',x_2')) \\
 &amp;\Rightarrow &amp;  \lambda x_1'+(1-\lambda)0 + f(\lambda x_2'+(1-\lambda)x_2'') &gt;  \lambda (x_1'+f(x_2'))+(1-\lambda)(0 + f(x_2''))\\
  &amp;\Rightarrow &amp;  \lambda x_1'+(1-\lambda)0 + f(\lambda x_2'+(1-\lambda)x_2'') &gt;  \lambda x_1'+(1-\lambda)0 + \lambda f(x_2')+(1-\lambda)f(x_2'') \\
    &amp;\Rightarrow &amp;  f(\lambda x_2'+(1-\lambda)x_2'') &gt;   \lambda f(x_2')+(1-\lambda)f(x_2'')\end{eqnarray*}
Therefore, $f(\cdot)$ is strict concave.