Minimum again?

Consider $f(x) = {(x + \frac{1}{x})}^{9}$ and we will see if this is convex by using the double derivative test.

$f'(x) = 9({(x + \frac{1}{x})}^{8})(1 -\frac{1}{{x}^{2}})$

$f''(x) = 72({(x + \frac{1}{x})}^{7})(1 - \frac{1}{{x}^{2}}) + 9(\frac{2}{{x}^{3}})({(x + \frac{1}{x})}^{8})$

Now we can see that f''(0) > 0, for positive x.

Hence, f(x) is convex and we can use Jensen's inequality to find the minimum.

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

$\geq 3(f(\frac{1}{3}))$

$\geq 3({(\frac{1}{3} + 3)}^{9})$

$\geq \frac{{10}^{9}}{{3}^{8}}$

Consider that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \geq 9 (use AM-HM)$ Notice that by adding $a+b+c$ , we have $(a+\frac{1}{a}) +( b+\frac{1}{b}) + (c+\frac{1}{c}) \geq 10$ Let $(a+\frac{1}{a}) = x, (b+\frac{1}{b}) = y, (c+\frac{1}{c}) = z$ Thus, we have $x+y+z \geq 10$ And by means, we have to search the minimum value of $x^9 +y^9 +z^9$ . Since we only have one equation and it is the only inequality, we have to work a little backwards to use Cauchy-Schwarz Inequality. By a little scratch, we have to use the power of 8 to use the Cauchy-Schwarz Inequality ( $8 = \frac{9}{2} + \frac{7}{2}$ ). Hence, we have

$(x^8 +y^8 +z^8)^2 \leq (x^9+y^9+z^9)(x^7+y^7+z^7) ... (1)$

But this inequality won't solve enough, so let us try make the powers of 7 using $\frac{8}{2} +\frac{6}{2}$ . Now, we have $(x^7+y^7+z^7)^2 \leq (x^8+y^8+z^8)(x^6+y^6+z^6)$ $\Rightarrow \frac{(x^7+y^7+z^7)^4}{(x^6+y^6+z^6)^2} \leq (x^8+y^8+z^8)^2 ... (2)$

Trailing the inequalities $(2)\leq (1)$ and simplication, we have $(x^7+y^7+z^7)^3 \leq (x^9+y^9+z^9)(x^6+y^6+z^6)^2$

Using Cauchy-Schwarz Inequality and this technique again and again, finally the inequality will be $(x+y+z)^9 \leq (x^9+y^9+z^9)(3^8)$

Since we know that $x+y+z \geq 10$ or $(x+y+z)^9 \geq 10^9$ , we just trail the inequalities to be $10^9 \leq (x^9+y^9+z^9)(3^8)$ $(x^9+y^9+z^9) \geq \frac{10^9}{3^8}$ Finally, $w=10, x=9, y=3, z=8 \Rightarrow w+x+y+z = 10+9+3+8=30$

Generalization For any positive reals of $x_1, x_2, ..., x_m$ , the inequality holds that $(\sum_{i=1}^m x_i)^n \leq m^{n-1}(\sum_{i=1}^m x_i^n)$ for any $n \in \mathbb{N}$

Generalization 2 (MORE GENERALIZED) For any positive reals of $x_1, x_2, ..., x_m$ , the inequality holds that $(\sum_{i=1}^m x_i^a)^{n-a+1} \leq (\sum_{i=1}^m x_i^n)(\sum_{i=1}^m x_i^{a-1})^{n-a}$ for $0\leq a\leq n$ $n,a \in \mathbb{N}$

Do you think that the Generalization 2 also holds for $n,a \in \mathbb{Z}$ ?

By using AM-GM

$\frac { a+b+c }{ 3 } \ge \frac { 3 }{ \frac { 1 }{ a } +\frac { 1 }{ b } +\frac { 1 }{ c } } \\ \frac { 1 }{ 3 } \ge \frac { 3 }{ \frac { 1 }{ a } +\frac { 1 }{ b } +\frac { 1 }{ c } } \\ \frac { 1 }{ a } +\frac { 1 }{ b } +\frac { 1 }{ c } \ge 9$

The minimum value can only be achieved when $a=b=c=\frac{1}{3}$ . ATQ

$3{ \left( 3+\frac { 1 }{ 3 } \right) }^{ 9 }\\ =\boxed{\frac{10^{9}}{3^{8}}}$