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

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

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

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

Log in to reply

Nice solution!

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.

Log in to reply

Nice!!!

Log in to reply

Pls give solutions

Log in to reply

It'd be too soon to give a solution now

Log in to reply