Let $a,b,c\ge0$ and $a+b+c=3.$ Prove that $(3a-bc)(3b-ac)(3c-ab)\le8.$ (O. Rudenko)
Problem
Source: Kyiv mathematical festival 2021
Tags: Kyiv mathematical festival, inequalities, algebra
spartacle
09.05.2021 04:06
It is equivalent to
$$\prod_{\text{cyc}}(a^2+ab+ca-bc) \le 8$$$$\iff \prod_{\text{cyc}} ((a+b)^2+(c+a)^2-(b+c)^2) \le 64$$
By Schur, the quantity is on the LHS is at most $$\prod_{\text{cyc}}(b+c)^2$$which is clearly at most $64$ by AM-GM.
VMF-er
09.05.2021 04:17
rogue wrote: Let $a,b,c\ge0$ and $a+b+c=3.$ Prove that $(3a-bc)(3b-ac)(3c-ab)\le8.$ (O. Rudenko)
Denote $bc+ca+ab=3(1-t^{2}),abc=r\ \ (0\leq t\leq 1)$. The inequality is equivalent to
$$f(r)=r^{2}-18(t^{2}+5)r+(81t^{2}-162t+89)\geq 0.$$Easy to see that $f(r)$ decreasing (because $r\leq 1$). Thus
$$f(r)\geq f((1-t)^{2}(1+2t))=t^{2}[(t-1)^{2}(t^{2}-10t+15)-4(t-1)+2]\geq 0.$$Equality occurs if and only if $a=b=c=1$. $\square$
snakeaid
13.05.2021 00:10
Obviously, it suffices to prove the inequality when all the factors in LHS are positive. By AM-GM $$\frac{(3a-bc)+(3b-ca)}{2}=\frac{(a+b)(a+b+c)-bc-ca}{2}=\frac{(a+b)^2}{2}\geq \sqrt{(3a-bc)(3b-ca)}$$So, taking the cyclic product, $$(3a-bc)(3b-ca)(3c-ab) \leq \frac{\Big((a+b)(b+c)(c+a)\Big)^2}{8}$$Now it suffices to prove that $$(a+b)(b+c)(c+a)\leq 8$$Which follows by AM-GM $$(a+b)(b+c)(c+a)\leq \Bigg(\frac{(a+b)+(b+c)+(c+a)}{3}\Bigg)^3=8 \; \blacksquare$$
sqing
24.05.2021 05:35
Let $a,b,c\ge0$ and $a+b+c=3.$ Prove that $$(a^2+a-bc)(b^2+b-ca)(c^2+c-ab)\le8 $$Let $a,b,c\ge0$ and $a^2+b^2+c^2=3.$ Prove that $$(3a^3-bc)(3b^3-ca)(3c^3-ab)\le8 $$Let $a,b,c\ge0$ and $a^2+b^2+c^2+abc=4.$ Prove that $$(3a^2-bc)(3b^2-ca)(3c^2-ab)\le8 $$Let $a,b,c>0$ such that $ a+b+c=1 .$ prove that $$ (a-bc)(b-ca)(c-ab)\le8(abc)^2 $$https://artofproblemsolving.com/community/c6h323597p1735789 $$ (ab + ac + bc)^3 - abc(a + b + c)^3 = (a^2-bc)(b^2-ca)(c^2-ab)$$ #2 $$\prod_{\text{cyc}} ((a+b)^2+(c+a)^2-(b+c)^2) \le \prod_{\text{cyc}}(b+c)^2$$https://artofproblemsolving.com/community/c6h598883p3554420 snakeaid wrote: Obviously, it suffices to prove the inequality when all the factors in LHS are positive. By AM-GM $$\frac{(3a-bc)+(3b-ca)}{2}=\frac{(a+b)(a+b+c)-bc-ca}{2}=\frac{(a+b)^2}{2}\geq \sqrt{(3a-bc)(3b-ca)}$$So, taking the cyclic product, $$(3a-bc)(3b-ca)(3c-ab) \leq \frac{\Big((a+b)(b+c)(c+a)\Big)^2}{8}$$Now it suffices to prove that $$(a+b)(b+c)(c+a)\leq 8$$Which follows by AM-GM $$(a+b)(b+c)(c+a)\leq \Bigg(\frac{(a+b)+(b+c)+(c+a)}{3}\Bigg)^3=8 \; \blacksquare$$ If $3a-bc<0,3b-ca<0$ then $$\frac{(3a-bc)+(3b-ca)}{2}<\sqrt{(3a-bc)(3b-ca)}$$
sqing
07.06.2023 03:55
Let $a,b,c\ge0$ and $a+b+c=3.$ Prove that $$ 8\geq (3a-bc)(3b-ca)(3c-ab)\geq -\frac{729}{16} $$$$ 8\geq (a^2+a-bc)(b^2+b-ca)(c^2+c-ab)\geq -\frac{2025}{64} $$Let $a,b,c\ge0$ and $a^2+b^2+c^2+abc=4.$ Prove that $$8\geq(3a^2-bc)(3b^2-ca)(3c^2-ab)\geq -72 $$
RenheMiResembleRice
26.08.2023 11:05
VMF-er wrote: rogue wrote: Let $a,b,c\ge0$ and $a+b+c=3.$ Prove that $(3a-bc)(3b-ac)(3c-ab)\le8.$ (O. Rudenko)
Denote $bc+ca+ab=3(1-t^{2}),abc=r\ \ (0\leq t\leq 1)$. The inequality is equivalent to
$$f(r)=r^{2}-18(t^{2}+5)r+(81t^{2}-162t+89)\geq 0.$$Easy to see that $f(r)$ decreasing (because $r\leq 1$). Thus
$$f(r)\geq f((1-t)^{2}(1+2t))=t^{2}[(t-1)^{2}(t^{2}-10t+15)-4(t-1)+2]\geq 0.$$Equality occurs if and only if $a=b=c=1$. $\square$
How can you denote bc+ca+ab equals this and abc equals that?