\[\large \dfrac{a^2+1}{4b^2}+\dfrac{b^2+1}{4c^2}+\dfrac{c^2+1}{4a^2}\geq\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\]

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

\[\large \dfrac{a^2+1}{4b^2}+\dfrac{b^2+1}{4c^2}+\dfrac{c^2+1}{4a^2}\geq\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\]

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

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TopNewest\[\text{By AM-GM we have:}\\ \color{Red}{\left(\frac{a²}{4b²}+\frac{1}{4a²}\right)}+\color{Blue}{\left(\frac{b²}{4c²}+\frac{1}{4b²}\right)}+\color{Cyan}{\left(\frac{c²}{4a²}+\frac{1}{4c²}\right)}\geq \color{Red}{\frac{1}{2b}}+\color{Blue}{\frac{1}{2c}}+\color{Cyan}{\frac{1}{2a}}\\ \text{So we have to prove that}\\ \frac{1}{2a}+\frac{1}{2b}+\frac{1}{2c}\geq \frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\\ \text{WLOG}\; a\geq b\geq c.\;\text{Let}\; f(a,b,c)=\frac{1}{2a}+\frac{1}{2b}+\frac{1}{2c}-\frac{1}{a+b}-\frac{1}{b+c}-\frac{1}{c+a}\\ ,\text{then we have}\; f(a,b,c)\geq f(a,b,b),\text{the easy proof of which is left to the reader}\\ \text{So now we have to prove that}\\ \begin{align} \frac{1}{2a}+\frac{1}{b}-\frac{2}{a+b}-\frac{1}{2b}&\geq 0\\ \implies \frac{1}{2a}+\frac{1}{2b}&\geq \frac{2}{a+b}\\ \frac{1}{a}+\frac{1}{b}&\geq\frac{4}{a+b}\\ \frac{a+b}{ab}&\geq\frac{4}{a+b}\\ \implies (a+b)²&\geq 4ab\\ \implies a²-2ab+b²=(a-b)² &\geq 0 \end{align}\\ \text{Which is obvious.Hence proved.}\] – Abdur Rehman Zahid · 8 months, 2 weeks ago

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

By applying AM-HM to, \[ \dfrac{\dfrac{1}{2a} + \dfrac{1}{2b}}{2} \geq \frac{2}{2a + 2b} = \frac{1}{a + b} \]

Adding up similar inequalities, we get the final inequality. – Ameya Daigavane · 8 months, 2 weeks ago

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– Abdur Rehman Zahid · 8 months, 2 weeks ago

Nice!!!Log in to reply

Pls give solutions – Harmeetsingh Chugga · 8 months, 3 weeks ago

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It'd be too soon to give a solution now – Gurīdo Cuong · 8 months, 3 weeks ago

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