microsoft_office_word wrote: Let $a,b,c \ge 0$ and $s \ge 5$. Prove that $s(a^2+b^2+c^2) \le s(s-3)(a^3+b^3+c^3)+1$ sure it is corect?

microsoft_office_word wrote: Let $a,b,c \ge 0$ and $s \ge 5$. Prove that $s(a^2+b^2+c^2) \le s(s-3)(a^3+b^3+c^3)+1$ Let $a^2+b^2+c^2=3t^2$. Using Power-Mean inequality, we get: $$a^3+b^3+c^3 \geq 3t^3$$Therefore, it is enough to check: $$f(t) = 3s(s-3)t^3+1-3st^2 \geq 0$$Calculating the derivative, we get: $$f'(t)=9s(s-3)t^2-6st$$Therefore, $f$ is minimal at: $$t_{0}=\frac{2}{3(s-3)}$$We have: $$f(t_0) = \frac{9 s^2 - 58 s + 81}{9 (s - 3)^2} \geq 0$$which is true for $s \geq 5$. (In fact, it is true for $s\geq \frac{29 + 4 \sqrt{7}}{9}$) The proof is complete.

Using same method we can prove, that $f(t)=s(s-3)t^3-st^2+\frac{5}{27} \geq 0 $ for $s \geq 5$

microsoft_office_word wrote: Let $a,b,c \ge 0$ and $s \ge 5$. Prove that $s(a^2+b^2+c^2) \le s(s-3)(a^3+b^3+c^3)+1$ Where is schur ?

Let $a,b,c \ge 0 .$ For $\lambda \ge\frac{29 + 4 \sqrt{7}}{9},$ prove that $$\lambda (a^2+b^2+c^2) \le\lambda (\lambda-3)(a^3+b^3+c^3)+1$$

teomihai wrote: microsoft_office_word wrote: Let $a,b,c \ge 0$ and $s \ge 5$. Prove that $s(a^2+b^2+c^2) \le s(s-3)(a^3+b^3+c^3)+1$ sure it is corect? My bad,now it is corect,i am very sorry

delete this

Let $a, b, c$ be $\frac{x}{x+y+z}, \frac{y}{x+y+z}$ and $\frac{z}{x+y+z}$ respectively. The inequality becomes $$ s \frac{x^2+y^2+z^2}{(x+y+z)^2} \le 3(s-3)\frac{x^3+y^3+z^3}{(x+y+z)^3} + 1$$which can be rewritten as: $$ 0 \le (s-5)(2\sum_{cyc} x^3 - \sum_{sym}x^2y) + 2(\sum_{cyc}x^3-\sum_{sym}x^2y+3xyz)$$ $2\sum_{cyc} x^3 \ge \sum_{sym}x^2y$ can be proved using the rearrangement inequality, and $$\sum_{cyc}x^3-\sum_{sym}x^2y+3xyz = \sum_{cyc}x(x-y)(x-z) \ge 0$$Equality holds for $a=b=c=\frac{1}{3}$ or $s=5, a=b=\frac{1}{2}, c=0$ and similar.

microsoft_office_word wrote: Let $a,b,c \ge 0$ and $s \ge 5$. Prove that $s(a^2+b^2+c^2) \le 3(s-3)(a^3+b^3+c^3)+1$ He forgot to mention that $a+b+c = 1$.

Let $a,b,c \ge 0$ and $a+b+c = 1 .$ Prove that $$5(a^2+b^2+c^2) \le 6(a^3+b^3+c^3)+1$$$$\implies$$Let $a,b,c \ge 0$ and $a+b+c = 1 .$ For $\lambda \ge 5,$ prove that $$\lambda (a^2+b^2+c^2) \le 3 (\lambda-3)(a^3+b^3+c^3)+1$$

Homogenize so it suffices to prove $$s(a^3+b^3+c^3)\leq (3s-9)(a^3+b^3+c^3)+(a+b+c)^3.$$Simplifying, the inequality turns into $$(2s-8)(a^3+b^3+c^3)+(3-s)\sum_{\text{sym}}a^2b+6abc\geq 0.$$From Schur we know $a^3+b^3+c^3+3abc\geq \sum_{\text{sym}}a^2b$ so it suffices to prove a weaker $$(2s-8)(a^3+b^3+c^3)+(3-s)(a^3+b^3+c^3+3abc)+6abc \geq 0\implies (a^3+b^3+c^3)(s-5)\geq (3s-15)abc,$$which is obvious from AM-GM.

\begin{align*} f(s) &= s(a^2+b^2+c^2)-3(s-3)(a^3+b^3+c^3) \\ &= s(a^2+b^2+(1-a-b)^2)-3(s-3)((a+b)(a^2-ab+b^2)+c^3) \\ &= s(2a^2+2ab+2b^2-2a-2b+1)-3(s-3)((a+b)(a^2-ab+b^2)+c^3) \\ &= s(\frac{3}{2}(a+b)^2+\frac{1}{2}(a-b)^2-2(a+b)+1)-3(s-3)((a+b)(\frac{3}{4}(a-b)^2+\frac{1}{4}(a+b)^2)+c^3) \\ &= s(\frac{3}{2}(1-c)^2+\frac{1}{2}(a-b)^2-2(1-c)+1)-3(s-3)((1-c)(\frac{3}{4}(a-b)^2+\frac{1}{4}(1-c)^2)+c^3) \\ &= \frac{-9s+27}{4}c^3+\frac{-3s+27}{4}c^2+\frac{5s-27}{4}c+\frac{-s+9}{4}+(\frac{9s-27}{4}c+\frac{-7s+27}{4})(a-b)^2. \end{align*}When $s \geq 5$, $\frac{9s-27}{4} > 0$. When $c \leq \frac{7s-27}{9s-27}$, $\frac{9s-27}{4}c+\frac{-7s+27}{4} \leq 0$ (obviously $\frac{4}{9} \leq \frac{7s-27}{9s-27} < \frac{7}{9}$). So \begin{align*} f(s) \leq \frac{-9s+27}{4}c^3+\frac{-3s+27}{4}c^2+\frac{5s-27}{4}c+\frac{-s+9}{4} &= (-\frac{9}{4}c^3-\frac{3}{4}c^2+\frac{5}{4}c-\frac{1}{4})s+\frac{27}{4}c^3+\frac{27}{4}c^2-\frac{27}{4}c+\frac{9}{4} \\ &= -\frac{1}{4}(3c-1)^2(c+1)s+\frac{27}{4}c^3+\frac{27}{4}c^2-\frac{27}{4}c+\frac{9}{4}. \end{align*}We can see that $-\frac{1}{4}(3c-1)^2(c+1) \leq 0$. So $f(s) \leq f(5) = -\frac{9}{2}c^3+3c^2-\frac{1}{2}c+1 = -\frac{1}{2}c(3c-1)^2+1 \leq 1$. When $c > \frac{7s-27}{9s-27}$, $\frac{9s-27}{4}c+\frac{-7s+27}{4} > 0$. We can easily know that $a-b \leq 1-c$, $b-a \leq 1-c$ ($a,b \geq 0$). So $(b-a)^2 \leq (1-c)^2$. \begin{align*} f(s) \leq \frac{-9s+27}{4}c^3+\frac{-3s+27}{4}c^2+\frac{5s-27}{4}c+\frac{-s+9}{4}+(\frac{9s-27}{4}c+\frac{-7s+27}{4})(1-c)^2 &= (-7s+27)c^2+(7s-27)c-2s+9 \\ &= (-7c^2+7c-2)s+27c^2-27c+9 \\ &= (-7(c-\frac{1}{2})^2-\frac{1}{4})s+27c^2-27c+9. \end{align*}We can see that $-7(c-\frac{1}{2})^2-\frac{1}{4} < 0$. So $f(s) \leq f(5) = -8c^2+8c-1 = -8(c-\frac{1}{2})^2+1 \leq 1$.

Let $a, b, c\geq 0$ and $a+b+c=1$. For $\lambda\geq 5$, prove that $$(\lambda-3)(a^3+b^3+c^3)-\lambda (a^2+b^2+c^2)+1\geq 0$$Clearly, the last expression is linear in $w^3$. Thus, it suffices to prove the inequality in two cases: First case. $c=0$ Our inequality is single-variable in $t=ab\leq \tfrac{1}{4}$, and it’s not difficult to prove! Second case. $b=c=t$ and $a=1-2t,$ where $t\leq \tfrac{1}{2}$ The desired inequality is equivalent to $$-2(3t-1)^2 ((\lambda-3)t-\lambda+4)=(-18\lambda+54)t^3+(30\lambda-108)t^2+(-14\lambda+54)t+(2\lambda-8)\geq 0$$Thus, the problem is reduced to proving $f(t)=(\lambda-3)t-\lambda+4\leq 0$. Clearly, $f$ is linear in $t$ and $\lambda-3>0$, thus it suffices to prove that $f(\tfrac{1}{2})\leq 0$, which is equivalent to $\lambda\geq 5$. $\blacksquare$

AlgebraFC wrote: Homogenize so it suffices to prove $$s(a^3+b^3+c^3)\leq (3s-9)(a^3+b^3+c^3)+(a+b+c)^3.$$Simplifying, the inequality turns into $$(2s-8)(a^3+b^3+c^3)+(3-s)\sum_{\text{sym}}a^2b+6abc\geq 0.$$From Schur we know $a^3+b^3+c^3+3abc\geq \sum_{\text{sym}}a^2b$ so it suffices to prove a weaker $$(2s-8)(a^3+b^3+c^3)+(3-s)(a^3+b^3+c^3+3abc)+6abc \geq 0\implies (a^3+b^3+c^3)(s-5)\geq (3s-15)abc,$$which is obvious from AM-GM. It is a nice weaker inequality,your first identity transformation is wonderful

AlgebraFC wrote: Homogenize so it suffices to prove $$s(a^3+b^3+c^3)\leq (3s-9)(a^3+b^3+c^3)+(a+b+c)^3.$$Simplifying, the inequality turns into $$(2s-8)(a^3+b^3+c^3)+(3-s)\sum_{\text{sym}}a^2b+6abc\geq 0.$$From Schur we know $a^3+b^3+c^3+3abc\geq \sum_{\text{sym}}a^2b$ so it suffices to prove a weaker $$(2s-8)(a^3+b^3+c^3)+(3-s)(a^3+b^3+c^3+3abc)+6abc \geq 0\implies (a^3+b^3+c^3)(s-5)\geq (3s-15)abc,$$which is obvious from AM-GM. Aaaa...I really don't know how to data-in formulas like you. In fact,(2s-8)(a^3+b^3+c^3)+(3-s)(a^2b+ab^2+b^2c+bc^2+c^2a+ca^2)+6abc>=0 ===>(2s-10)(a^3+b^3+c^3)+(5-s)(a^2b+ab^2+b^2c+bc^2+c^2a+ca^2)>=0 <===>2(a^3+b^3+c^3)>=a^2b+ab^2+b^2c+bc^2+c^2a+ca^2 p=a+b+c,q=ab+bc+ca,r=abc <===>2(p^3-3pq+3r)>=pq-3r <===>2+9r>=7q It is obvious because2p^3+9r>=7pq(p^3>=3pq,p^3+9r>=4pq)

Good hunting

microsoft_office_word wrote: Let $a,b,c \ge 0$ such that $a+b+c=1$ and $s \ge 5$. Prove that $s(a^2+b^2+c^2) \le 3(s-3)(a^3+b^3+c^3)+1$ \[\displaystyle 1+3(s-3)(a^3+b^3+c^3) - s(a^2+b^2+c^2) = \sum \left[(a+1)(s-5)+2a\right](a-b)(a-c) \geqslant 0.\]

microsoft_office_word wrote: Let $a,b,c \ge 0$ such that $a+b+c=1$ and $s \ge 5$. Prove that $s(a^2+b^2+c^2) \le 3(s-3)(a^3+b^3+c^3)+1$ First of all thiw is a linear fuction from $s$ so it is enought to prove it for $s=5$. Let $p=a+b+c$,$q=ab+bc+ca$ and $r=abc$ then we want to prove that $p^3+9r>=4pq$ which is true