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Let a, b, c >0

 

Prove that \(\frac{a^3}{b^2(5a+2b)}+\frac{b^3}{c^2(5b+2c)} + \frac{c^3}{a^2(5c+2a)} \geq \frac{3}{7}\).

 

I'm currently dying: 

 

Here is my thought process:

 

By the Cauchy-Schwarz inequality,

\(∑\limits_{cycl}a^3b^2(5a+2b)∑\limits_{cycl}[ab^2(5a+2b)]≥(a^2+b^2+c^2)^2.\)

Thus, \(∑\limits_{cycl}a^3b^2(5a+2b)≥(a^2+b^2+c^2)^2∑\limits_{cycl}[ab^2(5a+2b)]≥\frac{3}{7}.\)

 

I have no clue from where to go from here, if somesone could help me, it would be great.

 Feb 5, 2019
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\(\text{To prove:} \displaystyle \sum_{cyc(a,b,c)} \dfrac{a^3}{b^2 (5a+2b)} \geq \dfrac{3}{7}\\ \text{By Cauchy-Schwarz inequality: } \displaystyle \sum_{cyc(a,b,c)} \dfrac{a^3}{b^2 (5a+2b)} \sum_{cyc(a,b,c)}ab^2 (5a+2b) \geq (a^2+b^2+c^2)^2\\ \text{If we can show that } \displaystyle \sum_{cyc(a,b,c)} \dfrac{a^3}{b^2 (5a+2b)} \geq \dfrac{(a^2+b^2+c^2)^2}{\displaystyle\sum_{cyc(a,b,c)}ab^2 (5a+2b)} \geq \dfrac{3}{7}\text{, then we are done.}\\ \text{That means we now need to prove } \displaystyle 3 \sum_{cyc(a,b,c)} ab^2 (5a+2b) \leq 7(a^2+b^2+c^2)^2.\\ \text{We can easily see that: }3 \displaystyle\sum_{cyc(a,b,c)} ab^2 (5a+2b) = 15 \sum_{cyc(a,b,c)} a^2 b^2 + 6\sum_{cyc(a,b,c)}ab^3\\ \text{By AM-GM inequality: } \displaystyle3\sum_{cyc(a,b,c)} ab^2 (5a+2b) \geq 3\left(15 \left(a^{\frac{4}{3}}b^{\frac{4}{3}}c^{\frac{4}{3}}\right) + 6\left(a^{\frac{4}{3}}b^{\frac{4}{3}}c^{\frac{4}{3}}\right)\right) = 63 a^{\frac{4}{3}}b^{\frac{4}{3}}c^{\frac{4}{3}}\\ \text{In other words: }\displaystyle3\sum_{cyc(a,b,c)} ab^2 (5a + 2b) \geq 63a^{\frac{4}{3}}b^{\frac{4}{3}}c^{\frac{4}{3}}\\ \text{Which means: } 63a^{\frac{4}{3}}b^{\frac{4}{3}}c^{\frac{4}{3}} \leq 7(a^2+b^2+c^2)^2\text{, which is true.} \)

I don't know if this is correct or not. This is too challenging for a beginner in proving inequalities...

 Feb 7, 2019

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