Solving Algebraic Inequality with $n$ Positive Real Numbers

[lfdahl]

Given $n$ positive real numbers: $x_1,x_2,...,x_n$.

Show, that:

\[\frac{x_1^2}{x_2}+\frac{x_2^2}{x_3}+ ... + \frac{x_{n-1}^2}{x_n}+\frac{x_n^2}{x_1} \geq x_1+x_2+...+x_n\]

 

[Opalg]

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]

 

[lfdahl]

Good job, Opalg! Thankyou for your participation!

Your solution was exactly, what I had in mind!

 

[lfdahl]

An alternative solution can be found here:

Spoiler

By the AM-GM inequality, we have:

\[\frac{x_i^2}{x_{i+1}}+x_{i+1} \geq 2x_i \\\\ \Rightarrow \frac{x_1^2}{x_2}+\frac{x_2^2}{x_3}+...+\frac{x_n^2}{x_1}+x_1+x_2+...+x_n \geq 2\left ( x_1+x_2+...+x_n \right )\\\\ \Rightarrow \frac{x_1^2}{x_2}+\frac{x_2^2}{x_3}+...+\frac{x_n^2}{x_1} \geq x_1+x_2+...+x_n\]

 

[solakis1]

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]

 

[lfdahl]

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!