# Math Help - Inequality cubic-radical

Hello : is the positifs reales numbers :
prove :

2. What you probably mean is that if 'a' and 'b' are positive real numbers then prove that $b\geq 3\sqrt[3]{ab}-2\sqrt{a}$.

Solution

Let $x=\sqrt{a}$.Let $y=b$

Since $x$ and $y$ are positive real numbers,therefore

$\frac{x+x+y}{3}\geq \sqrt[3]{x^2y}$ since $A.M > G.M$

$2\sqrt{a}+b\geq 3\sqrt[3]{ab}$

$b\geq 3\sqrt[3]{ab}-2\sqrt{a}$.