We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
+1
145
1
avatar+79 

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
 #1
avatar+7685 
+2

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

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

 Feb 7, 2019

15 Online Users

avatar
avatar