Use am _ gm on 1st 3 terms

Urvs wrote: Use am _ gm on 1st 3 terms & use square root mean inequality on last u get the result. $\frac{a+b+c}{\sqrt{a^2+b^2+c^2}} \geq \sqrt{3} <=> \sum ab \geq \sum a^2$, which is false

Urvs wrote: Use am _ gm on 1st 3 terms & use square root mean inequality on last u get the result. it's not that easy!

augustin_p wrote: Urvs wrote: Use am _ gm on 1st 3 terms & use square root mean inequality on last u get the result. $\frac{a+b+c}{\sqrt{a^2+b^2+c^2}} \geq \sqrt{3} <=> \sum ab \geq \sum a^2$, which is false Yes i noted it.

We will show the following stronger inequality: $$\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+\sqrt{3}\frac{a+b+c}{\sqrt{a^2+b^2+c^2}} \geq 6.$$Proof: By Cauchy-Schwarz $$\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\geq \frac{(a+b+c)^2}{ab+bc+ca}$$and by AM-GM $$\frac{(a+b+c)^2}{ab+bc+ca}+\sqrt{3}\frac{a+b+c}{\sqrt{a^2+b^2+c^2}} \geq 2\sqrt[4]{3\frac{(a+b+c)^6}{(a^2+b^2+c^2)(ab+bc+ca)^2}}.$$Now, note that by AM-GM \begin{align*} (a+b+c)^2&=(a^2+b^2+c^2)+2(ab+bc+ca) \\ &\geq 3\sqrt[3]{(a^2+b^2+c^2)(ab+bc+ca)^2} \end{align*}and we're done.

splendid stronger inequality !

I was 2 minutes faster here

WolfusA wrote: I was 2 minutes faster here no problem, you are a very good and passionate settler of beautiful inequalities!

Let $ a$,$ b$,$ c$ be positive real numbers. Prove that$$\frac{a^2+b^2+c^2}{ab+bc+ca}+\frac{a+b+c}{\sqrt{a^2+b^2+c^2}} \geq 1+\sqrt{3}$$Let $ a$,$ b$,$ c$ be positive real numbers. Prove that $$\sqrt{\frac{a}{b}+\frac{b}{c}+\frac{c}{a}}+\sqrt{\frac{ab+bc+ca}{a^2+b^2+c^2}}\geq 1+\sqrt{3}$$

augustin_p wrote: Show that for any positive real numbers $a$, $b$, $c$ the following inequality takes place $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+\frac{a+b+c}{\sqrt{a^2+b^2+c^2}} \geq 3+\sqrt{3}$

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