is equivalent to $6+p^2+q^2+r^2 \ge 6max(p,q,r) \ when p+q+r=0 \\ let \ p=max(p,q,r) ,p \ge 0 \ q^2+r^2 \ge \frac 1 2 (q+r)^2 =\frac 1 2 p^2 \\ 6+p^2+q^2+r^2 \ge 6+\frac 3 2 p^2 \ge 6p $

forfun wrote: is equivalent to $6+p^2+q^2+r^2 \ge 6max(p,q,r) \ when p+q+r=0 \\ let \ p=max(p,q,r) ,p \ge 0 \ q^2+r^2 \ge \frac 1 2 (q+r)^2 =\frac 1 2 p^2 \\ 6+p^2+q^2+r^2 \ge 6+\frac 3 2 p^2 \ge 6p $ can you explain more your solution pleease

forfun let $p=a-b, q=b-c, r=c-a$. It's a nice solution btw

nsato wrote: Let $a$, $b$, $c$ be real numbers. Prove that \[ab + bc + ca + \max\{|a - b|, |b - c|, |c - a|\} \le 1 + \frac{1}{3} (a + b + c)^2.\] Define $f(a,b,c)=1 + \frac{1}{3} (a + b + c)^2-(ab + bc + ca + \max\{|a - b|, |b - c|, |c - a|\})$ Easy to see that $f(a,b,c)=f(a+t,b+t,c+t)$ $\forall t\in\mathbb{R}$ So $f(a,b,c)=f(0,b-a,c-a)$ and the rest is easy to verified

nsato wrote: Let $a$, $b$, $c$ be real numbers. Prove that \[ab + bc + ca + \max\{|a - b|, |b - c|, |c - a|\} \le 1 + \frac{1}{3} (a + b + c)^2.\] WLOG we can suppose $a\ge b\ge c$, so we need to prove: $a-c \leq 1+\frac{1}{6}\left( (a-b)^2+(b-c)^2+(c-a)^2 \right)$ Using Cauchy-Schwarz, we have: $(a-b)^2+(b-c)^2\ge \frac{1}{2}(a-c)^2$, so it remains to prove: $a-c \leq 1+\frac{(a-c)^2}{4} \Longleftrightarrow \left( \frac{a-c}{2}-1 \right)^2 \ge 0$ Equality holds when $a-b=b-c=1$ and all permutations. $\blacksquare$

nsato wrote: Let $a$, $b$, $c$ be real numbers. Prove that \[ab + bc + ca + \max\{|a - b|, |b - c|, |c - a|\} \le 1 + \frac{1}{3} (a + b + c)^2.\] $ \Leftrightarrow 4\left ( \left ( a+b+c \right )^{2}-3\left ( ab+bc+ca \right ) \right )+12\geq 12\max\{|a - b|, |b - c|, |c - a|\}\\\Leftrightarrow T=2\left ( \left ( a-b \right )^{2}+\left ( b-c \right )^{2}+\left ( c-a \right )^{2} \right )+12\geq 12\max\{|a - b|, |b - c|, |c - a|\}\\however ~~ 2\left ( \left ( a-b \right )^{2}+\left ( b-c \right )^{2}+\left ( c-a \right )^{2} \right )\geq 3\left ( a-b \right )^{2}\\~~and~~2\left ( \left ( a-b \right )^{2}+\left ( b-c \right )^{2}+\left ( c-a \right )^{2} \right )\geq 3\left ( b-c \right )^{2}\\ and~~2\left ( \left ( a-b \right )^{2}+\left ( b-c \right )^{2}+\left ( c-a \right )^{2} \right )\geq 3\left ( c-a \right )^{2}\\\Rightarrow T\geq 3\left ( \left ( a-b \right )^{2} +4\right )\geq 3\times 2 \sqrt{( a-b)^{2} \times 4}=12\left | a-b \right |\\and~~T\geq 3\left ( \left ( b-c \right )^{2} +4\right )\geq 3\times 2 \sqrt{( b-c)^{2} \times 4}=12\left | b-c \right | \\and~~T\geq 3\left ( \left ( c-a \right )^{2} +4\right )\geq 3\times 2 \sqrt{( c-a)^{2} \times 4}=12\left | c-a \right |\\finnally ~~T\geq 12\max\{|a - b|, |b - c|, |c - a|\}~~\blacksquare $