Proving an expression to be negative I have this expression which I want to prove that it's negative, given some constraints on it's variables. I've tried to maximise this function using Lagrange multipliers  hoping to find a maximum which is negative valued, but arrived to a dead end. 
The expression is as follows :  
such that $a,b$ are positive real numbers
$x+y \le 1$
and $x,y\ge0$
 A: Well, $x+y \le 1 \iff x \le 1-y$
$$(ax+by)^2-a^2x-b^2y = a^2x(x-1)+b^2y(y-1) + 2abxy \le a^2(1-y)((1-y)-1)+b^2y(y-1)+2aby(1-y)=y(y-1)(a-b)^2 \le 0$$
since $x,y\ge0$ and $x+y \le 1$ giving $y \le 1$
A: By the Cauchy-Schwarz inequality, we have
$$(ax+by)^2=(a\sqrt{x} \cdot \sqrt x+b\sqrt{y} \cdot \sqrt y)^2 \leqslant (a^2x+b^2y)(x+y) \leqslant (a^2x+b^2y).$$
So
$$(ax+by)^2 - (a^2x+b^2y) \leqslant 0.$$
Done
A: When $x=0$ or $y=0$ the inequality holds:
$$a^2x^2-a^2x=a^2x(x-1)\le 0\hbox{ as }0\le x\le 1$$
Let $k=x+y,\,ku=x,\,kv=y, u+v=1$, then the expression rewrites to
$$k^2(au+b(1-u))^2-a^2ku-b^2k(1-u)=$$
$$=\frac14(2 a k u - a - 2 b k u + 2 b k - b)^2 - \frac14(a^2 - 4 a b k + 2 a b + b^2)$$
Since the thing is a parabola in terms of $u$ unless $a=b$ it can me maximized only on the interval ends, so we have to check only $u=0$, $u=1$.
$$=b^2k(k-1)\le 0\hbox{ at }u=0$$
$$=a^2k(k-1)\le 0\hbox{ at }u=1$$
The one case left $a=b$ then $2aku-2bku$ vanishes in the expression above like for $u=0$ and we have the same $$=b^2k(k-1)\le 0$$
