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Inequality

\(a,b,c\) are positive real numbers and \(abc=1\). Prove that

\(\frac{1}{2+a} + \frac{1}{2+b} + \frac{1}{2+c} \geq \frac{1}{1+a+b} + \frac{1}{1+b+c} + \frac{1}{1+c+a}\)

Note by Fahim Shahriar Shakkhor
3 years, 3 months ago

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\(\dfrac{1}{2+a}+\dfrac{1}{2+b}+\dfrac{1}{2+c} \geq \dfrac{1}{1+a+b}+\dfrac{1}{1+b+c}+\dfrac{1}{1+c+a} \\ \Rightarrow \dfrac{(2+a)(2+b)+(2+b)(2+c)+(2+a)(2+c)}{(2+a)(2+b)(2+c)} \geq \dfrac{(1+a+b)(1+b+c)+(1+a+b)(1+a+c)+(1+a+c)(1+b+c)}{(1+a+b)(1+a+c)(1+b+c)} \\ \Rightarrow \dfrac{ (4+2a+2b+ab)+(4+2b+2c+bc)+(4+2a+2c+ac)}{8+4a+4b+4c+2ab+2ac+2bc+abc} \geq \dfrac{(1+a+2b+c+ab+bc+ac+b^2)+(1+2a+b+c+ab+bc+ac+a^2)+(1+a+b+2c+ab+bc+ac+c^2)}{2(a+b+c)+3(ab+bc+ca)+(a^2+b^2+c^2)+(a+b+c)(ab+bc+ca)} \\ \Rightarrow \dfrac{12+4(a+b+c)+(ab+bc+ca)}{9+4(a+b+c)+2(ab+bc+ca)} \geq \dfrac{3+4(a+b+c)+3(ab+bc+ca)+(a^2+b^2+c^2)}{2(a+b+c)+3(ab+bc+ca)+(a^2+b^2+c^2)+(a+b+c)(ab+bc+ca)} \)

Let \(x=a+b+c\) and \(y=ab+bc+ca \\ a^2+b^2+c^2=(a+b+c)^2-2(ab+bc+ca)=x^2-2y\)

Putting the values we get,

\(\displaystyle \dfrac{12+4x+y}{9+4x+2y} \geq \dfrac{3+4x+x^2+y}{2x+y+x^2+xy} \\\Rightarrow 3x^2y+xy^2+6xy-5x^2-y^2-24x-3y-27 \geq 0 \\\Rightarrow (\dfrac{5}{3}x^2y-5x^2)+(\dfrac{4}{3}x^2y-12x)+(\dfrac{xy^2}{3}-y^2)+(\dfrac{xy^2}{3}-3x) + (\dfrac{xy^2}{3}-3y) + (3xy-9x) + (3xy-27) \geq 0 \)

which is true because \(x,y\geq3\)

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