True or false?

Are the following statements true?

a) For a,b,c>0a,b,c> 0, (a3c+b3a+c3b)2≥3a2b2c2(ab+bc+ca)( a^{3}c + b^{3}a + c^{3}b)^{2} ≥ 3a^{2}b^{2}c^{2}(ab + bc + ca)

b) For k∈N∗k \in \mathbb N*, 1+1k2+1(k+1)2∈Q\sqrt{1 + \frac{1}{k^{2}} + \frac{1}{(k + 1)^{2}}} \in \mathbb Q

#Algebra

Note by Gabi Dobre
5 years, 3 months ago

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Comments

b)
        1+1k2+1(k+1)2\displaystyle\;\;\;\;\sqrt{1+\frac 1{k^2}+\frac 1{(k+1)^2}}
=k2(k+1)2+(k+1)2+k2k2(k+1)2\displaystyle=\sqrt{\frac{k^2(k+1)^2+(k+1)^2+k^2}{k^2(k+1)^2}}
=k4+2k3+3k2+2k+1k2(k+1)2\displaystyle=\sqrt{\frac{k^4+2k^3+3k^2+2k+1}{k^2(k+1)^2}}
=(k2+k+1)2k2(k+1)2\displaystyle=\sqrt{\frac{(k^2+k+1)^2}{k^2(k+1)^2}}
=k2+k+1k(k+1)\displaystyle=\frac{k^2+k+1}{k(k+1)}
∈Q\in \mathbb Q

展豪 張 - 5 years, 2 months ago

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I think you should try a). Its more difficult.

Gabi Dobre - 5 years, 2 months ago

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I know...so I try b) first, haha! I'm still thinking about it......

展豪 張 - 5 years, 2 months ago

Phew, another inequality, another wild ride!
 

Expanding the LHS,
(∑cyca3c)2=∑cyca6c2+2∑cyca4b3c (\sum\limits_{cyc}^{} a^3c)^2 = \sum\limits_{cyc}^{} a^6c^2 + 2\sum\limits_{cyc}^{} a^4b^3c

By AM - GM, a6c2+b6a2≥2a4b3c a^6c^2 + b^6a^2 \geq 2a^4b^3c
Adding similar inequalities, ∑cyca6c2≥∑cyca4b3c \sum\limits_{cyc}^{} a^6c^2 \geq \sum\limits_{cyc}^{} a^4b^3c

LHS≥3∑cyca4b3c≥3∑cyca3b3c2=RHS\text{LHS} \geq 3\sum\limits_{cyc}^{} a^4b^3c \geq 3\sum\limits_{cyc}^{} a^3b^3c^2 = \text{RHS}, by Muirhead.
If Muirhead is too uncomfortable/artificial for you, I'll show you another way to prove ∑cyca4b3c≥∑cyca3b3c2\sum\limits_{cyc}^{} a^4b^3c \geq \sum\limits_{cyc}^{} a^3b^3c^2
 

The inequality is equivalent to,
∑cyca2bc≥∑cycab \sum\limits_{cyc}^{} \dfrac{a^2b}{c} \geq \sum\limits_{cyc}^{} ab
Now, by AM-GM, 47a2bc+17c2ab+27b2ca≥ab \dfrac{4}{7}\dfrac{a^2b}{c} + \dfrac{1}{7}\dfrac{c^2a}{b} + \dfrac{2}{7}\dfrac{b^2c}{a} \geq ab
Adding similar inequalities for the other two terms, we are done.

Ameya Daigavane - 5 years, 2 months ago
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