Inequality

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

12+a+12+b+12+c≥11+a+b+11+b+c+11+c+a\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}

#Algebra #Inequality

Note by Fahim Shahriar Shakkhor
6 years, 11 months ago

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Comments

12+a+12+b+12+c≥11+a+b+11+b+c+11+c+a⇒(2+a)(2+b)+(2+b)(2+c)+(2+a)(2+c)(2+a)(2+b)(2+c)≥(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)⇒(4+2a+2b+ab)+(4+2b+2c+bc)+(4+2a+2c+ac)8+4a+4b+4c+2ab+2ac+2bc+abc≥(1+a+2b+c+ab+bc+ac+b2)+(1+2a+b+c+ab+bc+ac+a2)+(1+a+b+2c+ab+bc+ac+c2)2(a+b+c)+3(ab+bc+ca)+(a2+b2+c2)+(a+b+c)(ab+bc+ca)⇒12+4(a+b+c)+(ab+bc+ca)9+4(a+b+c)+2(ab+bc+ca)≥3+4(a+b+c)+3(ab+bc+ca)+(a2+b2+c2)2(a+b+c)+3(ab+bc+ca)+(a2+b2+c2)+(a+b+c)(ab+bc+ca)\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+cx=a+b+c and y=ab+bc+caa2+b2+c2=(a+b+c)2−2(ab+bc+ca)=x2−2yy=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,

12+4x+y9+4x+2y≥3+4x+x2+y2x+y+x2+xy⇒3x2y+xy2+6xy−5x2−y2−24x−3y−27≥0⇒(53x2y−5x2)+(43x2y−12x)+(xy23−y2)+(xy23−3x)+(xy23−3y)+(3xy−9x)+(3xy−27)≥0\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≥3x,y\geq3

Fahim Shahriar Shakkhor - 6 years, 8 months ago
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