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Inequality 53 (Vo Quoc Ba Can)

Problem:

Let $\displaystyle a,b,c>0$ such that $\displaystyle a+b+c=3$ and $\displaystyle a^2+b^2+c^2=4$. Find the minimum and the maximum value of the expression

$\displaystyle Q=\frac{a}{b}$.

Solution:

From the Cauchy – Schwarz Inequality we have that

$\displaystyle (a^2+b^2+c^2)\cdot\left[\frac{(a+b)^2}{a^2+b^2}+1\right]\geq (a+b+c)^2$,

or due to the hypothesis

$\displaystyle \frac{(a+b)^2}{a^2+b^2}\geq \frac{5}{4}$.

From this last Inequality, we acquire

$\displaystyle a^2+b^2\leq 8ab$,

Let us now divide with $\displaystyle ab$. Then we get that

$\displaystyle \frac{a}{b}+\frac{b}{a}\leq 8$.

Solving this Inequality, gives us

$\displaystyle 4-\sqrt{15}\leq \frac{a}{b}\leq 4+\sqrt{15}$.

Thus, $\displaystyle Q_{\min}=4-\sqrt{15}$ and $\displaystyle Q_{\max}=4+\sqrt{15}$, Q.E.D.