nimueh wrote: For all positive real numbers $a,b,c$ with $a+b+c=3$, show that $a^3b+b^3c+c^3a+9\geq 4(ab+bc+ca)$ solution $$\sum a^3b+9=\sum \frac{a^3b^3}{b^2}+9\geq \frac{(\sum ab)^3}{(\sum a)^2}+9=\frac{(ab+ac+bc)^3}{9}+3+3+3\geq 3(ab+ac+bc)+3$$so we need to show $$ab+ac+bc\le 3$$obvious by $3(ab+ac+bc)\le (a+b+c)^2$

nimueh wrote: For all positive real numbers $a,b,c$ with $a+b+c=3$, show that $a^3b+b^3c+c^3a+9\geq 4(ab+bc+ca)$ Proof of Zhangyanzong : $$\sum a^3b\geq \sum (a^3b+b+b)+\sum ab\geq 4\sum ab$$

Alternative: Begin by noting via Cauchy-Schwarz that $(a^3b+b^3c+c^3a)(ab+bc+ca)\geq (a^2b+b^2c+c^2a)^2$, and also that, $(a^2b+b^2c+c^2a)(a+b+c)\geq (ab+bc+ca)^2$. Hence, we have the following chain of inequalities: $$ (a^3b+b^3c+c^3a)(ab+bc+ca)(a+b+c)^2\geq (a^2b+b^2c+c^2a)^2(a+b+c)^2\geq (ab+bc+ca)^4, $$and hence, we have, $$ a^3b+b^3c+c^3a \geq \frac{(ab+bc+ca)^3}{9}. $$Now, letting $t=ab+bc+ca$, it suffices to show that, $t^3/9 +9 \geq 4t$, under the condition that $t=ab+bc+ca\in (0,3]$ (simply notice that $9=(a+b+c)^2\geq 3(ab+bc+ca)$, and $ab+bc+ca$ can be made arbitrarily small [ignore this detail though, not important]). But this is obvious.

note that by AM GM $a^3b + 2b \ge 3ab$ adding similar inequalities and adding it to the following inequality we get the desired. $ a+b+c= 3 = \frac{(a + b+ c)^2}{3} \ge ab+bc+ca$

nimueh wrote: For all positive real numbers $a,b,c$ with $a+b+c=3$, show that $a^3b+b^3c+c^3a+9\geq 4(ab+bc+ca)$ ${a^3b+ab+2b}\geq{4ab}$ Similarly for others so we could substitute 6=2×(a+b+c) ${ab+bc+ca}\leq{3}$

Medjl wrote: nimueh wrote: For all positive real numbers $a,b,c$ with $a+b+c=3$, show that $a^3b+b^3c+c^3a+9\geq 4(ab+bc+ca)$ solution $$\sum a^3b+9=\sum \frac{a^3b^3}{b^2}+9\geq \frac{(\sum ab)^3}{(\sum a)^2}+9=\frac{(ab+ac+bc)^3}{9}+3+3+3\geq 3(ab+ac+bc)+3$$so we need to show $$ab+ac+bc\le 3$$obvious by $3(ab+ac+bc)\le (a+b+c)^2$ Which inequality is that? I got it by Holder, but it isn't that direct.

Duarti wrote: Which inequality is that? I got it by Holder, but it isn't that direct. It is literally a direct Holder. \begin{align*} \left(\sum_{cyc} ab\right)^3 = \left(\sum_{cyc} \frac{ab}{\sqrt[3]{b^2}} \cdot \sqrt[3]{b} \cdot \sqrt[3]{b} \right)^3 &\leq \left(\sum_{cyc} \frac{a^3b^3}{b^2} \right)\left(\sum_{cyc} b \right)\left(\sum_{cyc} b \right) \\ &=\left(\sum_{cyc} \frac{a^3b^3}{b^2} \right)\left(\sum_{cyc} a \right)^2. \end{align*}

Can we replace the 9 by 3(a+b+c) then homogenize?

By Holder $(a^3b+b^3c+c^3a)(b+c+a)^2\ge (ab+bc+ca)^3 \implies (a^3b+b^3c+c^3a) \ge (ab+bc+ca)^3/9$. $(ab+bc+ca)=x$, we have to prove $x^3/9+9 \ge 4x$ or $(x-3)(x^2+3x-27)\ge 0$ which is true since $9=(a+b+c)^2\ge 3(ab+bc+ca)$.

nimueh wrote: For all positive real numbers $a,b,c$ with $a+b+c=3$, show that $$a^3b+b^3c+c^3a+9\geq 4(ab+bc+ca).$$ Pretty easy for it's position.

We will use Tangent line trick Define $f(x) = x^3 - 4x$. We want to prove $bf(a) + cf(b) + af(c) \ge -9$. Note that $f''(x) = 6x$ so $f(x)$ is convex for $x > 0$. Now we have $f(x) \ge f(a) + f'(a)(x-a)$ so for $a = 1$ we have $f(x) \ge -3 + (1-x) = -2-x$ so $bf(a) + cf(b) + af(c) \ge -2b-ab-2c-bc-2a-ca = -6-ab-bc-ca$ so we need to prove $ab + bc + ca \le 3$ which is obvious since $9 = (a+b+c)^2 \ge 3(ab+bc+ca)$.

Here's a cuter approach: By Cauchy-Schwarz inequality and using $a+b+c=3$, we have $$ (a^3b+b^3c+c^3a)(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}) \geq (a+b+c)^2 \Longrightarrow a^3b+b^3c+c^3a \geq 3abc $$Thus we need to show $$ 3abc+9 \geq 4(ab+bc+ca) $$which is the same as (since $a+b+c=3$) $$ 9abc+(a+b+c)^3 \geq 4(a+b+c)(ab+bc+ca) $$Using the fact that $(a+b+c)^3=a^3+b^3+c^3+3(a+b)(a+c)(b+c)$, $(a+b+c)(ab+bc+ca) = (a+b)(a+c)(b+c)+abc$ and $(a+b)(a+c)(b+c)=a^2b+a^2c+b^2a+b^2c+c^2a+c^2b+2abc$ for all $a$, $b$, $c$, it suffices to show that $$ 3abc+a^3+b^3+c^3 \geq a^2b+a^2c+b^2a+b^2c+c^2a+c^2b $$which is Schur's inequality for $n = 1$. We are done.

Maybe mine is shorter besides @above thanks for bumping $a^3b+b+b \geq 3ab$ $b^3c+c+c \geq 3bc$ $c^3a+a+a \geq 3ac$ Then left turns this $3 \geq ab+bc+ca$ That's true because $(a+b+c)^2 \geq 3(ab+bc+ca)$

This Solution has typo, look at #24 By Muirhead, $a^3b+b^3c+c^3a+9\geq a^2b^2+b^2c^2+c^2a^2+9$ $\geq \frac{(ab+bc+ca)^2}{3}+9\geq 4(ab+bc+ca)$ $=> 9\geq (ab+bc+ca)\left(\frac{12-ab-bc-ca}{3}\right)$ That implies $27\geq (ab+bc+ca)[12-(ab+bc+ca)]$ Since $ab+bc+ca\leq \frac{9}{3}=3$, this is trivial. The proof completed

@above 1.Muirhead using only the symmetric inequalities 2.$ab+bc+ca \geq 3$ is wrong

ismayilzadei1387 wrote: @above 1.Muirhead using only the symmetric inequalities 2.$ab+bc+ca \geq 3$ is wrong As İsmail said solution had typo I tried to say less than. The solution continues like that. And this inequality is not symmetric. .

achen29 wrote: Can we replace the 9 by 3(a+b+c) then homogenize? I think homogenizing with $(a+b+c)^2$ is better.

Let $a,b,c >0 $ and $a+b+c=3 .$ Show that $$a^3b+b^3c+c^3a+2abc+7\geq 4(ab+bc+ca)$$

Generalization 1 Let $a,b,c,k\in \mathbf{R^+}$ and $k\geq 1$ such that $a+b+c=3$. Then prove the following $$a^kb+b^kc+c^ka+3k\geq (k+1)(ab+bc+ca)$$

Generalization 2 Let $a,b,c,d,k\in \mathbf{R^+}$ and $k\geq 1$ such that $a+b+c+d=4$. Then prove the following $$a^kb+b^kc+c^kd+d^ka+4k\geq (k+1)(ab+bc+cd+da)$$

Generalization 3 Let $a,b,c,d,k\in \mathbf{R^+}$ and $k\geq 1$ such that $a+b+c+d\leq 4$. Then prove the following $$a^kb+b^kc+c^kd+d^ka+k(a+b+c+d)\geq (k+1)(ab+bc+cd+da)$$

1-Solution $$a^{k}b+b^kc+c^ka+3k=a^kb+b^kc+c^ka+(a+b+c)$$$$=a^kb+(k-1)b+b^kc+(k-1)c+c^ka+(k-1)a+(a+b+c)=a^kb+\overbrace{b+\cdots+b}^{k-1}+b^kc+\overbrace{c+\cdots+c}^{k-1}+c^ka+\overbrace{a+\cdots+a}^{k-1}+(a+b+c)$$$$\overbrace{\geq}^{AM-GM} k(ab+bc+ca)+(a+b+c)\geq (k+1)(ab+bc+ca) $$ $a+b+c=3$ so $a+b+c\geq 3\geq ab+bc+ca$. That makes the last inequality true.