Prove $-\sqrt(x)$ is a convex function I need to prove it's strictly concave up by using the definition $f(tx + (1-t)y) < tf(x) + (1-t)f(y)$.
I'm stuck on plugging the values in then to show the inequality.
$-\sqrt { (tx+(1-t)y) } <t(-\sqrt { x } )+(1-t)(-\sqrt { y } )$
Similar question found in Is √x concave?
 A: Using only definition, without derivative:
We want to prove that for $t\in(0,1)$ and $x\neq y$, holds $$\sqrt{tx+(1-t)y}>t\sqrt{x}+(1-t)\sqrt{y}$$ (these two questions are equivalent, obviously $f$ is concave iff $-f$ is convex).
Since both sides positive, the inequality is equivalent to $${tx+(1-t)y}>t^2x+(1-t)^2y+2t(1-t)\sqrt{xy}$$ which can be rearranged to $$t(1-t)x+t(1-t)y>2t(1-t)\sqrt{xy}$$ which is equivalent to $$x+y>2\sqrt{xy}$$ or $$(\sqrt{x}-\sqrt{y})^2>0$$ but this one needs no proof.
A: It's enough to show that 
\begin{align}
    tx + (1-t)y &\ge (t \sqrt x + (1-t) \sqrt y)^2 \\
                &= t^2 x + (1-t)^2 y + 2t(1-t)\sqrt{xy}.
\end{align}
since then we could take the square root of both sides (preserving the inequality, since $\sqrt{\cdot}$ is increasing on the positive reals) and negate both sides (reversing the inequality, and getting the claim you want to show).
After staring at the algebraic expressions above for a few minutes, we spot $2 \sqrt{xy}$, which suggests using the AM-GM inequality. We have:
\begin{align}
    \frac12 x + \frac12 y &\ge \sqrt{xy} \\
    t(1-t)x + t(1-t)y &\ge 2t(1-t) \sqrt{xy} \\
    t^2 x + t(1-t)x + (1-t)^2y + t(1-t)y &\ge t^2 x + (1-t)^2 y + 2t(1-t)\sqrt{xy}. \\
    t x + (1-t)y &\ge (t \sqrt x + (1-t)\sqrt y)^2 \\
    \sqrt{t x + (1-t)y} &\ge t \sqrt x + (1-t) \sqrt y.
\end{align}

Alternatively, we notice that if we move everything to one side, the expression factors nicely. (This is actually equivalent to one possible proof of AM-GM.)
\begin{align}
    t(1-t) (\sqrt x - \sqrt y)^2 &\ge 0\\
    t(1-t) x + t(1-t) y - 2 t(1-t) \sqrt{xy} &\ge 0\\
    t(1-t) x + t(1-t) y &\ge 2 t(1-t) \sqrt{xy}
\end{align}
and from there you proceed as above.
