Funny. Main Claim. $\frac{5}{2} \le a + b + c \le 3$. Proof. Let $a + b + c = x$, now notice that the condition could be rewritten as \[ ab + bc + ca = 2(a + b + c)^2 - 5(a + b + c) = 2x^2 - 5x \]Furthermore, notice that \[ (a + b + c)^2 \ge 3(ab + bc + ca) \Rightarrow x^2 \ge 3(2x^2 - 5x) \Rightarrow x \le 3 \]and since $a,b,c \ge 0$, we have \[ 2x^2 - 5x = ab + bc + ca \ge 0 \Rightarrow x(2x - 5) \ge 0 \Rightarrow x \ge \frac{5}{2} \]since $x \ge 0$. Now, notice that we want to prove $7abc \le 8x^2 - 30x + 25$. Therefore, notice that \[ 3abc(a + b + c) \le (ab + bc + ca)^2 \Rightarrow abc \le \frac{x(2x - 5)^2}{3} \]Therefore, it suffices to prove that \[ 7x(2x - 5)^2 \le 3(8x^2 - 30x + 25) \]which could be rewritten as \[ (x - 3)(2x - 5)(14x - 5) \le 0 \]However, the last inequality is true since $x - 3 \le 0$, and $2x - 5 \ge 0, 14x - 5 \ge 0$. Equality holds for $(a,b,c) = (1,1,1)$.

GorgonMathDota wrote: Funny. Main Claim. $\frac{5}{2} \le a + b + c \le 3$. how about the case $a=b=c=0$ ?

puntre wrote: GorgonMathDota wrote: Funny. Main Claim. $\frac{5}{2} \le a + b + c \le 3$. how about the case $a=b=c=0$ ? I'm leaving out edge case since it is easy to deal with them.

DNCT1 wrote: Let $a,b,c$ are non-negative numbers such that $$2(a^2+b^2+c^2)+3(ab+bc+ca)=5(a+b+c)$$then prove that $4(a^2+b^2+c^2)+2(ab+bc+ca)+7abc\le 25$ Let $$a+b+c=p,bc+ca+ab=q,abc=r.$$$\Rightarrow 0\leq q=2p^{2}-5p\leq \frac{p^{2}}{3}.$ $\Rightarrow p\leq 3; p=0\vee p\geq \frac{5}{2}$ We need to prove that $$7r\leq 8p^{2}-30p+25.$$$$7r\leq \frac{7pq}{9}=\frac{7p^{2}(2p-5)}{9}=8p^{2}-30p+25+\frac{(2p-5)(7p-15)(p-3)}{9}\leq 8p^{2}-30p+25.$$Equality occurs if and only if $a=b=c=1$. $\square$ PS: Thanks @below, my mistake.

VMF-er wrote: DNCT1 wrote: Let $a,b,c$ are non-negative numbers such that $$2(a^2+b^2+c^2)+3(ab+bc+ca)=5(a+b+c)$$then prove that $4(a^2+b^2+c^2)+2(ab+bc+ca)+7abc\le 25$ Let $$a+b+c=p,bc+ca+ab=q,abc=r.$$$\Rightarrow q=2p^{2}-5p\leq \frac{p^{2}}{3}.$ $\Rightarrow p\leq 3.$ We need to prove that $$7r\leq 8p^{2}-30p+25.$$Applying Schur's inequality $$7r\leq \frac{7p(4q-p^{2})}{9}=\frac{7p^{2}(7p-20)}{9}=8p^{2}-30p+25+\frac{(p-3)(49p^{2}-65p+75)}{9}\leq 8p^{2}-30p+25.$$Equality occurs if and only if $a=b=c=1$. $\square$ Schur's inequality is $r \ge \frac{p(4q - p^2)}{9}$, right?

Is it allowed to use la grange multipliers on this exam?

Let $a+b+c=3u^2,ab+bc+ca=3v^2,abc=w^3$. We have $2(9u^2-6v^2)+9v^2=15u\iff 6u^2-5u=v^2$. Since $6u^2-5u=v^2\geq 0$ and $6u^2-5u=v^2\leq u^2,$ we have $\frac{5}{6}\leq u\leq 1$. \[4(9u^2-6v^2)+6v^2+7w^3\overset{?}{\leq}25\]\[90u-72u^2+7w^3=36u^2-18v^2+7w^3\overset{?}{\leq} 25\]We have \[72u^2-90u+25-7w^3\geq 72u^2-90u+25-(42u^3-35u^2)=(1-u)(42u^2-65u+25)\]\[(1-u)(42u^2-65u+25)=(1-u)(42(u-\frac{5}{7})(u-\frac{5}{6})\geq 0\]As desired.$\blacksquare$

DNCT1 wrote: Let $a,b,c$ are non-negative numbers such that $$2(a^2+b^2+c^2)+3(ab+bc+ca)=5(a+b+c)$$then prove that $4(a^2+b^2+c^2)+2(ab+bc+ca)+7abc\le 25$ Equality holds when $ a=b=c=1$ or $ a=\frac{5}{2},b=c=0. $ here