MHB Can the Root Function Solve Inequalities?

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The discussion centers on proving that for a quadratic function f(x) = ax^2 + bx + c, where a, b, and c are positive real numbers, the inequality involving the 1024th root of the product of function values at non-negative real numbers holds true. Participants acknowledge errors in previously suggested solutions, with one user thanking another for pointing out these mistakes. The conversation emphasizes the importance of clarity and accuracy in mathematical proofs. The thread highlights the collaborative nature of problem-solving in mathematical discussions. Overall, the focus remains on the validity of the inequality and the need for precise solutions.
lfdahl
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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 $x_1,x_2,…,x_{1024}$

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

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

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.
 
Last edited:
lfdahl said:
Suggested solution:

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.
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
 
kaliprasad said:
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)
 
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