×

# True or false?

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 $$k \in \mathbb N*$$, $\sqrt{1 + \frac{1}{k^{2}} + \frac{1}{(k + 1)^{2}}} \in \mathbb Q$

Note by Gabi Dobre
1 year, 5 months ago

Sort by:

Phew, 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. · 1 year, 4 months ago

b)
$$\displaystyle\;\;\;\;\sqrt{1+\frac 1{k^2}+\frac 1{(k+1)^2}}$$
$$\displaystyle=\sqrt{\frac{k^2(k+1)^2+(k+1)^2+k^2}{k^2(k+1)^2}}$$
$$\displaystyle=\sqrt{\frac{k^4+2k^3+3k^2+2k+1}{k^2(k+1)^2}}$$
$$\displaystyle=\sqrt{\frac{(k^2+k+1)^2}{k^2(k+1)^2}}$$
$$\displaystyle=\frac{k^2+k+1}{k(k+1)}$$
$$\in \mathbb Q$$ · 1 year, 5 months ago

I think you should try a). Its more difficult. · 1 year, 5 months ago