General Mathematics and Inequalities

Question

If $a+b+c=2,a>0,b>0,c>0$, then $$\sqrt{a^ab^bc^c}+\sqrt{a^bb^cc^a}+\sqrt{a^cb^ac^b}\leq K$$ What is the best value of $K$ here?

Answer 1

For $a=2$ and $b=c\rightarrow0^+$ we get a value $2$.

We'll prove that it's a maximal value.

Indeed, since $\frac{a}{2}+\frac{b}{2}+\frac{c}{2}=1$, by AM-GM we obtain: $$\sqrt{a^ab^bc^c}+\sqrt{a^bb^cc^a}+\sqrt{a^cb^ac^b}=a^{\frac{a}{2}}b^{\frac{b}{2}}c^{\frac{c}{2}}+a^{\frac{b}{2}}b^{\frac{c}{2}}c^{\frac{a}{2}}+a^{\frac{c}{2}}b^{\frac{a}{2}}c^{\frac{b}{2}}\leq$$ $$\leq\frac{a}{2}\cdot a+\frac{b}{2}\cdot b+\frac{c}{2}\cdot c+\frac{b}{2}\cdot a+\frac{c}{2}\cdot b+\frac{a}{2}\cdot c+\frac{c}{2}\cdot a+\frac{a}{2}\cdot b+\frac{b}{2}\cdot c=$$ $$=\frac{(a+b+c)^2}{2}=2.$$ Done!

Answer 2

$a^{a/2}b^{b/2}c^{c/2}≤(a^2+b^2+c^2)/2$ by weighted AM-GM Use the same inequality on the other two terms and add the three inequalities to get $√a^ab^bc^c+√a^bb^cc^a+√a^cb^ac^b≤(a+b+c)^2/2=2$ with equality iff $a=b=c=2/3$