Inequality with quadratic and maximum Let $x,y$ be real numbers in $[0,1]$ such that $$x \ge y \geq \max\left\{x^2, \frac{2x}{3}, \frac{1+x}{4}, \frac{1}{3}\right\}.$$ Is it true that $x^2+2y \ge 2x$?
In order to prove this, the term $x^2$ is currently on the "wrong" side, between the condition and the needed inequality. Using a linear combination of the conditions $y\geq \dots$ is not sufficient.

nope not true...
let $x = 0.6, y=0.4, \max\left\{(0.6)^2, \frac{2*0.6}{3}, \frac{1+0.6}{4}, \frac{1}{3}\right\}=\max\left\{0.36, 0.4, 0.4, 0.33\right\} =0.4\\$
$\therefore x \geq y \geq \max\left\{x^2, \frac{2x}{3}, \frac{1+x}{4}, \frac{1}{3}\right\}$
holds when x = 0.6 and y = 0.4. Now check if $x^2+2y \geq 2x \implies 1.16 \geq 1.2$ which is false! Hence it is not true that $x^2+2y \geq 2x$
