An inequality exercise

a2+14b2+b2+14c2+c2+14a2≥1a+b+1b+c+1c+a\large \dfrac{a^2+1}{4b^2}+\dfrac{b^2+1}{4c^2}+\dfrac{c^2+1}{4a^2}\geq\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}

Prove the inequality above for a,b,c>0a,b,c>0 .

#Algebra

Note by P C
4 years, 11 months ago

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Comments

Pls give solutions

Harmeetsingh Chugga - 4 years, 11 months ago

By AM-GM we have:(a²4b²+14a²)+(b²4c²+14b²)+(c²4a²+14c²)≥12b+12c+12aSo we have to prove that12a+12b+12c≥1a+b+1b+c+1c+aWLOG  a≥b≥c.  Let  f(a,b,c)=12a+12b+12c−1a+b−1b+c−1c+a,then we have  f(a,b,c)≥f(a,b,b),the easy proof of which is left to the readerSo now we have to prove that12a+1b−2a+b−12b≥0  ⟹  12a+12b≥2a+b1a+1b≥4a+ba+bab≥4a+b  ⟹  (a+b)²≥4ab  ⟹  a²−2ab+b²=(a−b)²≥0Which is obvious.Hence proved.\text{By AM-GM we have:}\\ \color{#D61F06}{\left(\frac{a²}{4b²}+\frac{1}{4a²}\right)}+\color{#3D99F6}{\left(\frac{b²}{4c²}+\frac{1}{4b²}\right)}+\color{cyan}{\left(\frac{c²}{4a²}+\frac{1}{4c²}\right)}\geq \color{#D61F06}{\frac{1}{2b}}+\color{#3D99F6}{\frac{1}{2c}}+\color{cyan}{\frac{1}{2a}}\\ \text{So we have to prove that}\\ \frac{1}{2a}+\frac{1}{2b}+\frac{1}{2c}\geq \frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\\ \text{WLOG}\; a\geq b\geq c.\;\text{Let}\; f(a,b,c)=\frac{1}{2a}+\frac{1}{2b}+\frac{1}{2c}-\frac{1}{a+b}-\frac{1}{b+c}-\frac{1}{c+a}\\ ,\text{then we have}\; f(a,b,c)\geq f(a,b,b),\text{the easy proof of which is left to the reader}\\ \text{So now we have to prove that}\\ \begin{aligned} \frac{1}{2a}+\frac{1}{b}-\frac{2}{a+b}-\frac{1}{2b}&\geq 0\\ \implies \frac{1}{2a}+\frac{1}{2b}&\geq \frac{2}{a+b}\\ \frac{1}{a}+\frac{1}{b}&\geq\frac{4}{a+b}\\ \frac{a+b}{ab}&\geq\frac{4}{a+b}\\ \implies (a+b)²&\geq 4ab\\ \implies a²-2ab+b²=(a-b)² &\geq 0 \end{aligned}\\ \text{Which is obvious.Hence proved.}

Abdur Rehman Zahid - 4 years, 11 months ago

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Nice solution!
You can actually get there faster from the second step, where you showed it was sufficient to show, 12a+12b+12c≥1a+b+1b+c+1c+a \frac{1}{2a} + \frac{1}{2b} + \frac{1}{2c} \geq \frac{1}{a + b} + \frac{1}{b + c} + \frac{1}{c + a}

By applying AM-HM to, 12a+12b2≥22a+2b=1a+b \dfrac{\dfrac{1}{2a} + \dfrac{1}{2b}}{2} \geq \frac{2}{2a + 2b} = \frac{1}{a + b}

Adding up similar inequalities, we get the final inequality.

Ameya Daigavane - 4 years, 11 months ago

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Nice!!!

Abdur Rehman Zahid - 4 years, 11 months ago

It'd be too soon to give a solution now

P C - 4 years, 11 months ago
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