Let a,b,c be real positive numbers and abc=1

[Math Tutor]

Let [tex]a,b[/tex] and [tex]c[/tex] be positive real numbers such that [tex]abc=1[/tex]. Prove the inequality: [tex](\frac{a}{1+ab})^2+(\frac{b}{1+bc})^2+(\frac{c}{1+ca})^2\geq \frac{3}{4}[/tex].

[Math Tutor]

Author: ins-

[Guest]

Spoiler: show

By the AM-GM inequality we know that
[tex](\frac{a}{1+ab})^2+(\frac{b}{1+bc})^2+(\frac{c}{1+ca})^2 \geq 3((\frac{a}{1+ab})^2(\frac{b}{1+bc})^2(\frac{c}{1+ca})^2)^{1/3}[/tex]
[tex]= 3(\frac{abc}{(1+ab)(1+bc)(1+ca)})^{2/3}[/tex]
[tex]= 3((1+ab)(1+bc)(1+ca))^{-2/3}[/tex]
[tex]= 3(1+ab+bc+ca+a^2bc+ab^2c+abc^2+a^2b^2c^2)^{-2/3}[/tex]
[tex]= 3(2+a+\frac{1}{a}+b+\frac{1}{b}+c+\frac{1}{c})^{-2/3}[/tex]
[tex]\geq 3(2+2+2+2)^{-2/3}[/tex]
[tex]= \frac{3}{4}[/tex]

R. Baber.

[Guest]

Apologies, my previous post was nonsense. The last inequality is false, I forgot about the negative sign in the exponent.

R. Baber.

[Guest]

A correct proof (hopefully):

Spoiler: show

Let [tex]x = a, y = 1, z = 1/b[/tex] which implies [tex]a = x/y, b = y/z, c = z/x[/tex]. This means
[tex](\frac{a}{1+ab})^2 + (\frac{b}{1+bc})^2 + (\frac{c}{1+ca})^2 = (\frac{xz}{yz+xy})^2 + (\frac{xy}{xz+yz})^2 + (\frac{yz}{xy+xz})^2[/tex].
Let [tex]p = xy+yz, q = xy+xz, r = yz+xz[/tex] this means the left hand side is
[tex]= (\frac{-p+q+r}{2p})^2 + (\frac{p+q-r}{2r})^2 + (\frac{p-q+r}{2q})^2[/tex]
[tex]\geq \frac{1}{12}(\frac{-p+q+r}{p} + \frac{p-q+r}{q} + \frac{p+q-r}{r})^2[/tex] (by Chebyshev's inequality)
[tex]= \frac{1}{12}(-3+(\frac{q}{p} + \frac{p}{q}) +(\frac{r}{p}+\frac{p}{r}) +(\frac{r}{q}+ \frac{q}{r}))^2[/tex]
[tex]\geq \frac{1}{12}(-3+2+2+2)^2[/tex] (by AM-GM)
[tex]=\frac{3}{4}[/tex]

R. Baber.

[MM]

Nice proof! Here is another one which beggins in the same way.

Spoiler: show

Let [tex]a=\frac{x}{y}[/tex],[tex]b=\frac{y}{z}[/tex] and [tex]c=\frac{z}{x}[/tex]. Now the inequality is [tex]\sum\frac{(xz)^2}{(x+z)^2y^2}\ge \frac{3}{4}[/tex]. Now use that [tex](x+z)^2\le2\left(x^2+z^2\right)[/tex] and we obtain [tex]\sum\frac{(xz)^2}{x^2y^2+z^2y^2}\ge \frac{3}{2}[/tex] which is Nesbitt's Inequality.