sqing wrote: Find the maximum value of the positive real number $k$ such that the inequality $$\frac{1}{kab+c^2} +\frac{1} {kbc+a^2} +\frac{1} {kca+b^2} \geq \frac{k+3}{a^2+b^2+c^2} $$holds for all positive real numbers $a,b,c$ such that $a^2+b^2+c^2=2(ab+bc+ca).$ Given a fixed $\lambda>1$, the numbers $a,b,c \geq 0$ satisfy $$a^2+b^2+c^2=\lambda (ab+bc+ca)> 0.$$Prove that $$\frac{{2\lambda + 1}}{{{a^2} + {b^2} + {c^2}}} \le \frac{1}{2 a b+c^2}+\frac{1}{2 b c+a^2}+\frac{1}{2 c a+b^2}$$$$ \leq \begin{cases}\frac{2 \lambda^2+\lambda}{2\left(a^2+b^2+c^2\right)} & : \textrm { if} \lambda \geq \frac{5}{2} \\ \frac{27 \lambda}{\left(-2 \lambda^2+7 \lambda+4\right)\left(a^2+b^2+c^2\right)} & : \textrm { if} 1<\lambda \leq \frac{5}{2}\end{cases} $$

Inequalities with beautiful solutions $2.113 $ Let a, b, c be real numbers such that $a^2+b^2+c^2=2(ab+bc+ca).$ Prove that $$\frac{|a-b|}{\sqrt{2ab+c^2}} +\frac{|b-c|} {\sqrt{2bc+a^2}} +\frac{|c-a|} {\sqrt{2ca+b^2}} \geq 2 $$

SerdarBozdag wrote:

How to claim $$\frac{(\sum (a-b)^2)^2}{\sum (c^2+2ab)(a-b)^2}=2$$

uvw works nicely: The inequality for $k=2$ is quadratic in $w$, but the expression we need to prove to be non-negative (after clearing denominators and bringing everything on one side) is divisible by $w$ after using $3u^2=4v^2$ (since the remainder after division by $w$ is a homogeneous polynomial in $u,v$ of degree $6$ and hence after using $3u^2=4v^2$ a constant times $u^6$, hence zero since we have the equality case $a=1, b=1, c=0$). Hence we only need to prove a linear inequality in $w$ so that it suffices to check the cases $c=0$ and $b=c$. If $c=0$, we have $(a-b)^2=0$ and hence w.l.o.g. $(1,1,0)$ which gives an equality case. If $b=c$, we have $a=4b$ and hence w.l.o.g. $(4,1,1)$ which gives the second equality case.

sqing wrote: Find the maximum value of the positive real number $k$ such that the inequality $$\frac{1}{kab+c^2} +\frac{1} {kbc+a^2} +\frac{1} {kca+b^2} \geq \frac{k+3}{a^2+b^2+c^2} $$holds for all positive real numbers $a,b,c$ such that $a^2+b^2+c^2=2(ab+bc+ca).$ $k=2:$

sqing wrote:

Inequalities with beautiful solutions $2.113 $ Let a, b, c be real numbers such that $a^2+b^2+c^2=2(ab+bc+ca).$ Prove that $$\frac{|a-b|}{\sqrt{2ab+c^2}} +\frac{|b-c|} {\sqrt{2bc+a^2}} +\frac{|c-a|} {\sqrt{2ca+b^2}} \geq 2 $$ $$\sum \frac{|a-b|}{\sqrt{2ab+c^2}}=\sum \frac{|a-b|}{\sqrt{(b-c)^2+(c-a)^2}}=\sum \sqrt{\frac{(a-b)^2}{(c-a)^2+(b-c)^2}} \ge 2$$

sqing wrote: Inequalities with beautiful solutions $2.113 $ Let a, b, c be real numbers such that $a^2+b^2+c^2=2(ab+bc+ca).$ Prove that $$\frac{|a-b|}{\sqrt{2ab+c^2}} +\frac{|b-c|} {\sqrt{2bc+a^2}} +\frac{|c-a|} {\sqrt{2ca+b^2}} \geq 2 $$ It was here