Prove that, if $x,y,z \in [1, \frac{4}{3}]$, then:
a)$x \sqrt{4-3z}+y \sqrt{4-3x}+z \sqrt{4-3y} \le 4$;
b)$xy \sqrt{4-3z}+xz \sqrt{4-3y}+yz \sqrt{4-3x} \le x^2+y^2+z^2$.
Ineq 2
For discussing Olympiad Level Algebra (and Inequality) problems
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