Are the following statements true?

a) For \(a,b,c> 0\), \[( a^{3}c + b^{3}a + c^{3}b)^{2} ≥ 3a^{2}b^{2}c^{2}(ab + bc + ca)\]

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

Are the following statements true?

a) For \(a,b,c> 0\), \[( a^{3}c + b^{3}a + c^{3}b)^{2} ≥ 3a^{2}b^{2}c^{2}(ab + bc + ca)\]

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

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TopNewestPhew, another inequality, another wild ride!

Expanding the LHS,

\( (\sum\limits_{cyc}^{} a^3c)^2 = \sum\limits_{cyc}^{} a^6c^2 + 2\sum\limits_{cyc}^{} a^4b^3c\)

By AM - GM, \( a^6c^2 + b^6a^2 \geq 2a^4b^3c \)

Adding similar inequalities, \( \sum\limits_{cyc}^{} a^6c^2 \geq \sum\limits_{cyc}^{} a^4b^3c \)

\(\text{LHS} \geq 3\sum\limits_{cyc}^{} a^4b^3c \geq 3\sum\limits_{cyc}^{} a^3b^3c^2 = \text{RHS}\), by Muirhead.

If Muirhead is too uncomfortable/artificial for you, I'll show you another way to prove \(\sum\limits_{cyc}^{} a^4b^3c \geq \sum\limits_{cyc}^{} a^3b^3c^2 \)

The inequality is equivalent to,

\( \sum\limits_{cyc}^{} \dfrac{a^2b}{c} \geq \sum\limits_{cyc}^{} ab \)

Now, by AM-GM, \( \dfrac{4}{7}\dfrac{a^2b}{c} + \dfrac{1}{7}\dfrac{c^2a}{b} + \dfrac{2}{7}\dfrac{b^2c}{a} \geq ab \)

Adding similar inequalities for the other two terms, we are done. – Ameya Daigavane · 1 year, 3 months ago

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b)

\(\displaystyle\;\;\;\;\sqrt{1+\frac 1{k^2}+\frac 1{(k+1)^2}}\)

\(\displaystyle=\sqrt{\frac{k^2(k+1)^2+(k+1)^2+k^2}{k^2(k+1)^2}}\)

\(\displaystyle=\sqrt{\frac{k^4+2k^3+3k^2+2k+1}{k^2(k+1)^2}}\)

\(\displaystyle=\sqrt{\frac{(k^2+k+1)^2}{k^2(k+1)^2}}\)

\(\displaystyle=\frac{k^2+k+1}{k(k+1)}\)

\(\in \mathbb Q\) – 展豪 張 · 1 year, 3 months ago

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– Gabi Dobre · 1 year, 3 months ago

I think you should try a). Its more difficult.Log in to reply

– 展豪 張 · 1 year, 3 months ago

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