Question

If $\frac{{\log }_{e}a}{b-c}=\frac{{\log }_{e}b}{c-a}=\frac{{\log }_{e}c}{a-b}$, then prove that $a^{b+c}+b^{c+a}+c^{a+b}\ge 3$.

Answer 1

Since $a>0,b>0$ & $c>0$
Given $\frac{{\log }_{e}a}{b-c}=\frac{{\log }_{e}b}{c-a}=\frac{{\log }_{e}c}{c-b}$

$\Rightarrow \frac{(b+c){\log }_{e}a}{b^2-c^2}=\frac{(c+a){\log }_{e}b}{c^2-a^2}=\frac{(a+b){\log }_{e}c}{a^2-b^2}=k\left(say\right)$

$\Rightarrow (b+c){\log }_{e}a=k(b^2-c^2)$ ....(1)

$(c+a){\log }_{e}b=k(c^2-a^2)$ ....(2)

$(a+b){\log }_{e}c=k(a^2-b^2)$ ......(3)

Adding these equations, we get

$(b+c){\log }_{e}a+(c+a){\log }_{e}b+(a+b){\log }_{e}c=k(b^2-c^2+c^2-a^2+a^2-b^2)$

$\Rightarrow (b+c){\log }_{e}a+(c+a){\log }_{e}b+(a+b){\log }_{e}c=0$
$\therefore {\log }_e{a}^{b+c}+{\log }_e{b}^{c+a}+{\log }_e{c}^{a+b}=0,[\because x\log a=\log a^x]$

$\Rightarrow {\log }_e(a^{b+c}.b^{c+a}.c^{a+b})=0,[\because \log a+\log b+\log c=\log (abc\left)\right]$

$\Rightarrow a^{b+c}.b^{c+a}.c^{a+b}=1$ ........(4)

Again$A.M.\ge G.M.$

$\frac{a^{b+c}+b^{c+a}+c^{a+b}}{3}\ge (a^{b+c}{b}^{c+a}{c}^{a+b}{)}^{1/3}=(1{)}^{1/3}=1$, using (4)

$\Rightarrow a^{b+c}+b^{c+a}+c^{a+b}\ge 3.$ $\left[Henceproved\right]$