Home » Uncategorized » Inequality 53 (Vo Quoc Ba Can)

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.

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