Question

Let a,b,c be positive real numbers such that a+b+c= 3. Show that a^b*b^c*c^a is less than equal to 1
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Answer 1

Since $x^2\geq0,$ we can have,

$\longrightarrow (a-b)^2+(b-c)^2+(c-a)^2\geq0$

because each term in LHS is non - negative, so will be the whole LHS. Equality holds true if and only if $a=b=c.$

Expanding the LHS, we get,

$\longrightarrow 2[(a^2+b^2+c^2)-(ab+bc+ca)]\geq0$

$\longrightarrow (a^2+b^2+c^2)-(ab+bc+ca)\geq0$

$\longrightarrow a^2+b^2+c^2\geq ab+bc+ca$

Adding $2(ab+bc+ca)$ to both sides,

$\longrightarrow a^2+b^2+c^2+2(ab+bc+ca)\geq ab+bc+ca+2(ab+bc+ca)$

$\longrightarrow(a+b+c)^2\geq3(ab+bc+ca)$

Given that $a+b+c=3.$

$\longrightarrow3^2\geq3(ab+bc+ca)$

$\longrightarrow 3(ab+bc+ca)\leq9$

Dividing by 9,

$\longrightarrow \dfrac{ab+bc+ca}{3}\leq1\quad\quad\dots(1)$

By AM - GM Inequality, we can have,

$\longrightarrow\dfrac{ab+bc+ca}{3}\geq\sqrt[3]{a^b\,b^c\,c^a}$

So (1) can become,

$\longrightarrow \sqrt[3]{a^b\,b^c\,c^a}\leq1$

$\longrightarrow\underline{\underline{a^b\,b^c\,c^a\leq1}}$

Hence Proved!