The powers of ka+bc, kc+ab, kb+ac are 1/2.

VicKmath7 wrote: $a, b, c$ are positive real numbers and $a+b+c=k$. Find the minimum value of $$\frac{b^2}{(ka+bc)^{\frac{1}{2}}}+\frac{c^2}{(kb+ac)^{\frac{1}{2}}}+\frac{a^2}{(kc+ab)^{\frac{1}{2}}} $$ LaTeXed

Thank you

Solution: By Titu's Lemma - \begin{align*} \sum _{cyc} \dfrac{a^2}{\sqrt{kc + ab}} & \geq \dfrac{(a+b+c)^2}{ \sum \sqrt {(c+a)(c+b)}} \\ &= \dfrac{k^2}{\sum \sqrt{ (c+a)(c+b)}} \end{align*} Now, $(c+a)(c+b) = (k-b)(k-a)$ By AM-GM, $$(k-a) + (k-b) \geq 2 \sqrt{(k-a)(k-b)}$$ So, \begin{align*} \dfrac{k^2}{\sum \sqrt{ (c+a)(c+b)} } &= \dfrac{2 k^2}{ \sum 2 \sqrt{ (k-b)(k-a)}} \\ & \geq \dfrac{2 k^2}{\sum (2k - b - a)} \\ &= \frac{2 k^2}{6k - 2k} \\ &= \boxed{ \frac{k}{2}} \end{align*} Equality is attained at $$a = b = c = \frac{k}{3}$$

I don't have solutions, but I think yes

Titu's lemma is Cauchy-Schwarz inequality, right?

VicKmath7 wrote: Titu's lemma is Cauchy-Schwarz inequality, right? Yes.

VicKmath7 wrote: a, b, c are positive real numbers and a+b+c=k. Find the minimum value of $ b^2/(ka+bc)^1/2+c^2/(kb+ac)^1/2+a^2/(kc+ab)^1/2 $ https://artofproblemsolving.com/community/c6h1316434p7073475: Let $a,b,c\geq 0$ and $a+b+c=k.$ Find the minimum value $M$ such that $M=\frac{b^2}{\sqrt{ka+bc}}+\frac{a^2}{\sqrt{kc+ab}}+\frac{c^2}{\sqrt{kb+ca}}.$