Still AM-GM

[dduclam]

If [tex]a,b,c>0[/tex] such that [tex]abc=1[/tex] then [tex]a+b+c\ge\frac2{1+a}+\frac2{1+b}+\frac2{1+c}[/tex]

[MM]

"pqr" "kills" it easy as some people say.
First of all we join the fractions. We obtain [tex]\frac{a^{2}b+a^{2}c+b^{2}a+b^{2}c+c^{2}a+c^{2}b+3abc+a^{2}bc+b^{2}ca+c^{2}ba+a^{2}+b^{2}+c^{2}-3a-3b-3c-6}{(1+a) (1+b)(1+c)}\ge 0[/tex]. Now it's enough to prove that the numerator is bigger or equal to 0. Let a+b+c=p and ab+ac+bc=q. Then using abc=1 the numerator is equal to [tex]pq+p^{2}-2q-2p-6\ge 0[/tex] => [tex](p-2)(p+q)\ge 6[/tex]. Using AM-GM we obtain [tex]a+b+c\ge 3\sqrt[3]{abc}=3[/tex] and [tex]ab+ac+bc\ge 3\sqrt[3]{(abc)^{2}}=3[/tex]. Therefore [tex](p-2)(p+q)\ge (3-2)(3+3)=6[/tex]. Done!

[JusTok]

The inequality is equivalent to:
[tex]\frac{a+b+c}{2}\ge 3 - \frac{a}{a+1} - \frac{b}{b+1} - \frac{c}{c+1}[/tex]
[tex]\frac{a+b+c-6}{2} \ge -(\frac{a}{a+1} + \frac{b}{b+1} + \frac{c}{c+1})[/tex]
[tex]\frac{a}{a+1} + \frac{b}{b+1} + \frac{c}{c+1}\ge \frac{6-(a+b+c)}{2}[/tex]
We use AM≥HM for the left side:
[tex]\frac{9}{\frac{a+1}{a}+\frac{b+1}{b}+\frac{c+1}{c}}\ge \frac{6-(a+b+c)}{2}[/tex]
[tex]\frac{9}{3+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}\ge \frac{6-(a+b+c)}{2}[/tex]
[tex]\frac{9}{3+\frac{9}{a+b+c}}\ge \frac{6-(a+b+c)}{2}[/tex]
[tex]\frac{3}{1+\frac{3}{a+b+c}}\ge \frac{6-(a+b+c)}{2}[/tex]
[tex]\frac{3}{\frac{3+a+b+c}{a+b+c}}\ge \frac{6-(a+b+c)}{2}[/tex]
[tex]\frac{3(a+b+c)}{3+a+b+c}\ge \frac{6-(a+b+c)}{2}[/tex]
[tex]6(a+b+c)\ge (6-(a+b+c))(3+(a+b+c)[/tex]
[tex]6(a+b+c)\ge 18+3(a+b+c)-(a+b+c)^2[/tex]
[tex](a+b+c)(6-3+a+b+c)\ge 18[/tex]
[tex]a+b+c \ge3[/tex]
[tex]3(6-3+3)\ge 18[/tex]
[tex]18 \ge 18[/tex]

[MM]

Using AM-GM we obtain [tex]x+1\ge 2\sqrt{x}[/tex]. After applying this to the denominator in the right side of the inequality we have to prove that [tex]a+b+c\ge \frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}}[/tex]. Now we obtain [tex](a+b+c)\sqrt{abc}\ge \sqrt{ab}+\sqrt{ac}+\sqrt{bc}\Leftrightarrow a+b+c\ge \sqrt{ab}+\sqrt{ac}+\sqrt{bc}\Leftrightarrow(\sqrt{a}-\sqrt{b})^{2}+(\sqrt{c}-\sqrt{a})^{2}+(\sqrt{b}-\sqrt{c})^{2}\ge 0[/tex], QED!