Problem 30

DIRECT APPLICATION OF POWER MEAN INEQUALITY

$\sqrt{\frac{a^2+b^2+c^2}{3}} \leq \sqrt[3]{\frac{a^3+b^3+c^3}{3}}$

$(\sqrt{\frac{a^2+b^2+c^2}{3}})^6 \leq (\sqrt[3]{\frac{a^3+b^3+c^3}{3}})^6$

$\frac{(a^2+b^2+c^2)^3}{27} \leq \frac{(a^3+b^3+c^3)^2}{9}$

$(a^2+b^2+c^2)^3 \leq \boxed{3}(a^3+b^3+c^3)^2$

So, our answer is $\boxed{3}$

Applying $A.M. \geq G.M.$ :-

$a^3 + b^3 + c^3 \geq 3 (abc)^{\frac{3}{3}}$

$$

$\displaystyle \Rightarrow (a^3 + b^3 + c^3)^2 \geq 9 (abc)^2$

$$

$\displaystyle \Rightarrow (a^3 + b^3 + c^3)^2_{min} = 9 (abc)^2 \ldots equation\{1\}$

$$

and

$$

$a^2 + b^2 + c^2 \geq 3 (abc)^{\frac{2}{3}}$

$$

$\displaystyle \Rightarrow (a^2 + b^2 + c^2)^{3} \geq 27 (abc)^2 \ldots equation\{2\}$

Equality occurs when $a=b=c$ in both the cases.

Now, we write the given inequality as:

$$

$K \times (a^3 + b^3 + c^3)^2 \geq (a^2 + b^2 + c^2)^3 \geq 27 (abc)^2 \ldots \text{ from equation{2}}$

$$

$\displaystyle \Rightarrow K \times (a^3 + b^3 + c^3)^2 \geq (a^2 + b^2 + c^2)^3 \geq 3\times (a^3 + b^3 + c^3)^2_{min} \ldots \text{ from equation{1}}$

$$

$\displaystyle \Rightarrow K \times (a^3 + b^3 + c^3)^2 \geq 3\times (a^3 + b^3 + c^3)^2_{min}$

$$

Equality occurs for $a=b=c$ , substituting:

$$

$K\times (3a^3)^2 = 3\times (9\times a^2 \times a^2 \times a^2)$

$$

$\displaystyle \Rightarrow K\times 9a^6 = 3\times 9 a^6$

$$

$\displaystyle \Rightarrow \boxed{ K= 3}$

We know A.M of mth powers $\ge$ mth powers of A.M.

$\frac{a^2+b^2+c^2}{3}\ge\frac{(a+b+c)^2}{9}$

or, $(a^2+b^2+c^2)^3\ge\frac{(a+b+c)^6}{81}$

Similarly, $(a^3+b^3+c^3)^2\ge\frac{(a+b+c)^6}{27}$

$K\ge\frac{\frac{(a+b+c)^6}{27}}{\frac{(a+b+c)^6}{81}}$

Therefore, $K\ge3$

We have four solution:Two solution using AM-GM and power mean inequality.And 2 another is Jensen and Holder.