08-03-09 Prove this inequality.

Another Inequality by AM GM alone.

\sqrt{\frac{a+b}{c}}+\sqrt{\frac{b+c}{a}}+\sqrt{\frac{c+a}{b}}\geq 3\sqrt{2}

Hint: AM GM inequality twice.

10 Answers

11
virang1 Jhaveri ·

a , b,c are in Ap or Gp?

62
Lokesh Verma ·

none..

they are just 3 numbers! (+ve)!

21
tapanmast Vora ·

try this guyzzzz

11
Subash ·

i think i got it

but one thing shouldnt the inequality be

>=3 cubeth root of 2

replyy soon

11
Subash ·

alrite im posting what i did

applying amgm for the first oneq\geq {2abc+ac(a+c)+bc(b+c)+ab(a+b)}/abc

RHS under power of 1/6

applying amgm to the terms like ac(a+c) we can replace them by 2(ac)3/2

\geq {2+2((ac)^(3/2)/abc+similar terms)}

again applying amgm inequality to the three other terms we get the result

in the second latex equation the term is (ac)3/2/abc

62
Lokesh Verma ·

could you write the whole thing more clearly...

I think you could be quite close or have already solved the problem

1
krish1092 ·

\sum_{cyc}\sqrt{\frac{a+b}{c}}\ge 3(\sqrt{\frac{(a+b)(b+c)(c+a)}{abc}})^{1/3} \\ \text{Now} \\ \frac{a+b}{2} \ge \sqrt{ab} \ [\text{prceeding similarly}] \\ \\ \Rightarrow (a+b)(b+c)(c+a) \ge 8abc \\ \sum_{cyc}\sqrt{\frac{a+b}{c}}\ge 3(\sqrt{(\frac{8abc}{abc}}))^{1/3} \\ \sum_{cyc}\sqrt{\frac{a+b}{c}}\ge 3\sqrt2

62
Lokesh Verma ·

Great work :)

1
voldy ·

I dunno where I'm going wrong ?

1
voldy ·

oh got it . nice work da krish long time :)

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