Rukevwe wrote: Let $a, b, c, \lambda$ be positive real numbers with $\lambda \geq 1/4$. Show that $$\frac{a}{\sqrt{b^2+\lambda bc+c^2}}+\frac{b}{\sqrt{c^2+\lambda ca+a^2}}+\frac{c}{\sqrt{a^2+\lambda ab+b^2}} \geq \frac{3}{\sqrt{\lambda +2}}.$$ Use Holder and Schur

Let's complete arqady's proposal. Denote the given object by $S$. Then by Holder, \[ S^2\left(\sum a\bigl(b^2+\lambda bc+c^2\bigr)\right)\ge (a+b+c)^3, \]yielding \[ S^2\ge \frac{(a+b+c)^3}{\sum ab^2 + 3\lambda abc}. \]We now show the last quantity is at least $9/(\lambda+2)$. To that end, it suffices to show \[ (\lambda+2)\left(\sum a^3 + 6abc + 3\sum ab^2\right)\ge 9\sum ab^2 + 27\lambda abc. \]By Schur's inequality, $\sum a(a-b)(a-c)\ge 0$, we have $\sum a^3+3abc \ge \sum a^2b$. With this, we have \[ (\lambda+2)\left(\sum a^3 + 6abc + 3\sum ab^2\right)\ge 4(\lambda+2)\sum ab^2 + 3(\lambda+2)abc. \]With this, we are left with proving \[ (4\lambda-1)\sum ab^2 \ge (4\lambda-1)6abc, \]which is trivial by AM-GM as $\lambda \ge 1/4$.

Rukevwe wrote: Let $a, b, c, \lambda$ be positive real numbers with $\lambda \geq 1/4$. Show that $$\frac{a}{\sqrt{b^2+\lambda bc+c^2}}+\frac{b}{\sqrt{c^2+\lambda ca+a^2}}+\frac{c}{\sqrt{a^2+\lambda ab+b^2}} \geq \frac{3}{\sqrt{\lambda +2}}.$$ it is one nice generalization for one classical problem.

Rukevwe wrote: Let $a, b, c, \lambda$ be positive real numbers with $\lambda \geq 1/4$. Show that $$\frac{a}{\sqrt{b^2+\lambda bc+c^2}}+\frac{b}{\sqrt{c^2+\lambda ca+a^2}}+\frac{c}{\sqrt{a^2+\lambda ab+b^2}} \geq \frac{3}{\sqrt{\lambda +2}}.$$ This one screams Law of Cosine.

harryhero wrote: Rukevwe wrote: Let $a, b, c, \lambda$ be positive real numbers with $\lambda \geq 1/4$. Show that $$\frac{a}{\sqrt{b^2+\lambda bc+c^2}}+\frac{b}{\sqrt{c^2+\lambda ca+a^2}}+\frac{c}{\sqrt{a^2+\lambda ab+b^2}} \geq \frac{3}{\sqrt{\lambda +2}}.$$ This one screams Law of Cosine. no it is not possible with cosinus law: $$c^2 = a^2 + b^2 - 2ab \cos \angle C$$

teomihai wrote: harryhero wrote: Rukevwe wrote: Let $a, b, c, \lambda$ be positive real numbers with $\lambda \geq 1/4$. Show that $$\frac{a}{\sqrt{b^2+\lambda bc+c^2}}+\frac{b}{\sqrt{c^2+\lambda ca+a^2}}+\frac{c}{\sqrt{a^2+\lambda ab+b^2}} \geq \frac{3}{\sqrt{\lambda +2}}.$$ This one screams Law of Cosine. no it is not possible with cosinus law: $$c^2 = a^2 + b^2 - 2ab \cos \angle C$$ Think about three triangles each with an angle that creates the $+ \lambda ab$ term. (But for $\lambda \ge 2,$ we'll have to do something a bit more clever.)

Let $a, b, c$ be positive real numbers. Prove that$$\frac{a}{\sqrt{b^2+\lambda bc+c^2}}+\frac{b}{\sqrt{c^2+\lambda ca+a^2}}+\frac{c}{\sqrt{a^2+\lambda ab+b^2}} \geq \frac{3}{\sqrt{\lambda +2}}$$Where $\lambda \ge1 .$ Let $a, b, c$ be positive real numbers. Show that $$\frac{a}{\sqrt{b^2+bc+c^2}}+\frac{b}{\sqrt{c^2+ca+a^2}}+\frac{c}{\sqrt{a^2+ab+b^2}} \geq \frac {a+b+c}{\sqrt{ab+bc+ca}} \geq \sqrt{3}$$

arqady wrote: Rukevwe wrote: Let $a, b, c, \lambda$ be positive real numbers with $\lambda \geq 1/4$. Show that $$\frac{a}{\sqrt{b^2+\lambda bc+c^2}}+\frac{b}{\sqrt{c^2+\lambda ca+a^2}}+\frac{c}{\sqrt{a^2+\lambda ab+b^2}} \geq \frac{3}{\sqrt{\lambda +2}}.$$ Use Holder and Schur very nice