Question

If $a,b,c\in +R$ , then the minimum value of $a^{\log b-\log c}+b^{\log c-\log a}+c^{\log a-\log b}$ is

• A. $3$
• B. $1$
• C. $9$
• D. $16$

Answer 1

Answer: (A)

$3$

We know that $A.M.\ge G.M.$

$⟹\frac{a^{\log b-\log c}+b^{\log c-\log a}+c^{\log a-\log b}}{3}\ge \sqrt[3]{3a^{\log b-\log c}\times b^{\log c-\log a}\times c^{\log a-\log b}}$

$⟹\frac{a^{\log b-\log c}+b^{\log c-\log a}+c^{\log a-\log b}}{3}\ge \sqrt[3]{3\frac{a^{\log b}}{a^{\log c}}\times \frac{b^{\log c}}{b^{\log a}}\times \frac{c^{\log a}}{c^{\log b}}}$

$⟹\frac{a^{\log b-\log c}+b^{\log c-\log a}+c^{\log a-\log b}}{3}\ge \sqrt[3]{3\frac{a^{\log b}}{a^{\log c}}\times \frac{b^{\log c}}{a^{\log b}}\times \frac{a^{\log c}}{b^{\log c}}}$ (since, $a^{logb}=b^{loga}$)

$⟹\frac{a^{\log b-\log c}+b^{\log c-\log a}+c^{\log a-\log b}}{3}\ge \sqrt[3]{31}$

$⟹a^{\log b-\log c}+b^{\log c-\log a}+c^{\log a-\log b}\ge 3$

Therefore, the minimum value of $a^{\log b-\log c}+b^{\log c-\log a}+c^{\log a-\log b}$ is $3$