Can the Root Function Solve Inequalities?

[lfdahl]

Suppose, that $f(x)=ax^2+bx+c$, where $a$,$b$ and $c$ are positive real numbers. Show, that for all non-negative real numbers $x_1,x_2,…,x_{1024}$

\[\sqrt[1024]{f(x_1)\cdot f(x_2)\cdot \cdot \cdot f(x_{1024})} \geq f\left ( \sqrt[1024]{x_1\cdot x_2\cdot \cdot \cdot x_{1024}} \right )\]

 

[lfdahl]

Hint:

Spoiler

It might be of use first to show, that
\[f(x)\cdot f(y)\geq \left (f\left ( \sqrt{xy} \right ) \right )^2\]

 

[lfdahl]

Suggested solution:

Spoiler

Let us show, that for all $x,y > 0$

\[f(x)\cdot f(y)\geq \left (f\left ( \sqrt{xy} \right ) \right )^2 \;\;\; (1)\]

Indeed,

\[f(x)\cdot f(y) - \left (f\left ( \sqrt{xy} \right ) \right )^2 = ab(x^2y+xy^2-2xy\sqrt{xy})+ac(x^2+y^2-2xy)+bc(x+y-2\sqrt{xy}) \\\\ =abxy(\sqrt{x}-\sqrt{y})^2+ac(x-y)^2+bc(\sqrt{x}-\sqrt{y})^2 \geq 0.\]

Now by $(1)$

\[f(x_1)f(x_2)...f(x_{1024})\geq \prod_{i=1,3,5,…}^{1023}\left ( f\left ( \sqrt{x_i\; x_{i+1}} \right ) \right )^2 \geq \prod_{i=1,5,9,…}^{1021}\left ( f\left ( \sqrt[4]{x_i\;x_{i+1}\; x_{i+2}\; x_{i+3}} \right ) \right )^4 \geq ...\geq \left ( f \left ( \sqrt[1024]{x_1x_2...x_{1024}} \right ) \right )^{1024}\]

The required inequality readily follows.

 

[kaliprasad]

Suggested solution:

Spoiler

Let us show, that for all $x,y > 0$

\[f(x)\cdot f(y)\geq \left (f\left ( \sqrt{xy} \right ) \right )^2 \;\;\; (1)\]

Indeed,

\[f(x)\cdot f(y) - \left (f\left ( \sqrt{xy} \right ) \right )^2 = ab(x^2y+xy^2-2xy\sqrt{xy})+ac(x^2+y^2-2xy)+bc(x+y-2\sqrt{xy}) \\\\ =abxy(\sqrt{x}-\sqrt{y})^2+ac(x-y)^2+bc(\sqrt{x}-\sqrt{y})^2 \geq 0.\]

Now by $(1)$

\[f(x_1)f(x_2)...f(x_{1024})\geq \prod_{i=1}^{1023}\left ( f\left ( \sqrt{x_i\; x_{i+1}} \right ) \right )^2 \geq \prod_{i=1}^{1021}\left ( f\left ( \sqrt{x_i\;x_{i+1}\; x_{i+2}\; x_{i+3}} \right ) \right )^2 \geq ...\geq f\left ( \left ( \sqrt{x_1x_2...x_{1024}} \right ) \right )^{1024}\]

The required inequality readily follows.

Spoiler

in my humble opinion in the 1st product i change by 2 in the 2nd by 4 so on (this is not depicted in expression) and the 1st power is 2 in second 4 so on

 

[lfdahl]

Spoiler

in my humble opinion in the 1st product i change by 2 in the 2nd by 4 so on (this is not depicted in expression) and the 1st power is 2 in second 4 so on

Thankyou, kaliprasad, for pointing out the errors in the suggested solution.(Handshake)
I´m sorry for having posted a solution with severe typos.(Sadface)