MHB Solving Algebraic Inequality with $n$ Positive Real Numbers

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The discussion focuses on proving the inequality involving $n$ positive real numbers, specifically showing that the sum of certain fractions is greater than or equal to the sum of the numbers themselves. The Cauchy–Schwarz inequality is applied to derive the result, leading to the conclusion that the inequality holds true. Alternative proofs are also presented, emphasizing different approaches to reach the same conclusion. The participants express appreciation for each other's contributions and solutions. The overall consensus is that the inequality is valid for all positive real numbers.
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Given $n$ positive real numbers: $x_1,x_2,...,x_n$.

Show, that:

\[\frac{x_1^2}{x_2}+\frac{x_2^2}{x_3}+ ... + \frac{x_{n-1}^2}{x_n}+\frac{x_n^2}{x_1} \geq x_1+x_2+...+x_n\]
 
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lfdahl said:
Given $n$ positive real numbers: $x_1,x_2,...,x_n$.

Show, that:

\[\frac{x_1^2}{x_2}+\frac{x_2^2}{x_3}+ ... + \frac{x_{n-1}^2}{x_n}+\frac{x_n^2}{x_1} \geq x_1+x_2+...+x_n\]
[sp]Apply the Cauchy–Schwarz inequality $|\mathbf{x.y}|^2 \leqslant \|\mathbf{x}\|^2\|\mathbf{y}\|^2$ to the vectors $$\mathbf{x} = \left(\frac{x_1}{\sqrt{x_2}},\frac{x_2}{\sqrt{x_3}},\ldots,\frac{x_{n-1}}{\sqrt{x_n}},\frac{x_n}{\sqrt{x_1}}\right), \ \mathbf{y} = \left(\sqrt{x_2},\sqrt{x_3},\ldots,\sqrt{x_n},\sqrt{x_1}\right),$$ getting $$(x_1+x_2+\ldots+x_n)^2 \leqslant \left(\frac{x_1^2}{x_2}+\frac{x_2^2}{x_3}+ \ldots + \frac{x_{n-1}^2}{x_n}+\frac{x_n^2}{x_1}\right)\left(x_2+x_3 + \ldots + x_n+x_1\right).$$ Then divide both sides by $x_1+x_2+\ldots+x_n$ to get the result.[/sp]
 
Good job, Opalg! Thankyou for your participation!

Your solution was exactly, what I had in mind!
 
Last edited:
An alternative solution can be found here:

By the AM-GM inequality, we have:

\[\frac{x_i^2}{x_{i+1}}+x_{i+1} \geq 2x_i \\\\ \Rightarrow \frac{x_1^2}{x_2}+\frac{x_2^2}{x_3}+...+\frac{x_n^2}{x_1}+x_1+x_2+...+x_n \geq 2\left ( x_1+x_2+...+x_n \right )\\\\ \Rightarrow \frac{x_1^2}{x_2}+\frac{x_2^2}{x_3}+...+\frac{x_n^2}{x_1} \geq x_1+x_2+...+x_n\]
 
We have the following high school proof

[sp]Given, $$x_1,x_2,...x_n$$ positive real Nos we have:
$$(\frac{x_1}{\sqrt x_2}X+\sqrt x_2)^2+...(\frac{x_{n-1}}{\sqrt x_n}+\sqrt x_n)^2+(\frac{x_n}{\sqrt x_1}+\sqrt x_1)^2\geq 0$$
,which is true for all n and for all X............1
..........OR...........
$$(\frac{{x_1}^2}{x_2}+...\frac{x_{n-1}^2}{x_n}+\frac{{x_n}^2}{x_1})X^2+2(x_1+x_2+...x_n)X+(x_1+x_2+...x_n)\geq 0$$

Put $$A=(\frac{{x_1}^2}{x_2}+...\frac{x_{n-1}^2}{x_n}+\frac{{x_n}^2}{x_1})$$
$$B= x_1+x_2+...x_n $$ and (1) becomes:

$$AX^2+2BX+B\geq 0$$............(2)
.........OR.......
$$A(X+\frac{B}{A})^2+\frac{AB-B^2}{A}\geq 0$$...............3

Now since its equivalent (1) is true for all X ,hence (3) will be true for all X and particularly for $$X=-\frac{B}{A}$$.

And (3) becomes :

$$AB-B^2\geq 0$$

.........OR.........

$$A\geq B$$

Which is the desired inequality

Note : the same type of proof i suggested for the thread "high school inequality 3" but in a kind of backwards working .There the initial inequality should be:

$$(a_1X+b_1)^2+...(a_nX+b_1)^2\geq 0$$ [/sp]
 
solakis said:
We have the following high school proof

[sp]Given, $$x_1,x_2,...x_n$$ positive real Nos we have:
$$(\frac{x_1}{\sqrt x_2}X+\sqrt x_2)^2+...(\frac{x_{n-1}}{\sqrt x_n}+\sqrt x_n)^2+(\frac{x_n}{\sqrt x_1}+\sqrt x_1)^2\geq 0$$
,which is true for all n and for all X............1
..........OR...........
$$(\frac{{x_1}^2}{x_2}+...\frac{x_{n-1}^2}{x_n}+\frac{{x_n}^2}{x_1})X^2+2(x_1+x_2+...x_n)X+(x_1+x_2+...x_n)\geq 0$$

Put $$A=(\frac{{x_1}^2}{x_2}+...\frac{x_{n-1}^2}{x_n}+\frac{{x_n}^2}{x_1})$$
$$B= x_1+x_2+...x_n $$ and (1) becomes:

$$AX^2+2BX+B\geq 0$$............(2)
.........OR.......
$$A(X+\frac{B}{A})^2+\frac{AB-B^2}{A}\geq 0$$...............3

Now since its equivalent (1) is true for all X ,hence (3) will be true for all X and particularly for $$X=-\frac{B}{A}$$.

And (3) becomes :

$$AB-B^2\geq 0$$

.........OR.........

$$A\geq B$$

Which is the desired inequality

Note : the same type of proof i suggested for the thread "high school inequality 3" but in a kind of backwards working .There the initial inequality should be:

$$(a_1X+b_1)^2+...(a_nX+b_1)^2\geq 0$$ [/sp]
Well done, solakis! Thankyou for your participation!
 
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