The inequality is equilavent to $$3\left(a^2+b^2\right)\geq 4abc^2+a^3+b^3$$Since $a^3+b^3=\left(a+b\right)\left(a^2+b^2\right)-ab(a+b)$, we have $$\Longleftrightarrow \left(3-a-b\right)\left(a^2+b^2\right)+ab(a+b)\geq 4abc^2$$by the problem condition, it boils down proving $$a^2c+b^2c+a^2b+ab^2\geq 4abc^2$$which is true by of course AM-GM $$a^2c+b^2c+a^2b+ab^2\geq 4ab\sqrt[4]{abc^2}\geq 4abc^2\Longleftrightarrow c\leq 1$$.

2 min solve by $s=a+b$ and $p=ab$. (By the way, $a\geq b \geq 1$ is not needed, only $c\leq 1$ suffices.) We have $2 \leq s < 3$ and want to prove $\frac{3(s^2-2p)}{p} \geq 4(3-s)^2 + \frac{s^3-3sp}{p}$, equivalent to $\frac{s^2(3-s)}{p} \geq 4s^2 - 27s + 42$. Since $p \leq \frac{s^2}{4}$ (equivalent to $(a-b)^2 \geq 0$), it suffices to have $4(3-s) \geq 4s^2-27s+42$. This is equivalent to $4s^2 - 23s + 30 \leq 0$, i.e. $(s-2)(2s-15) \leq 0$, which holds for the given interval for $s$.

$(a+b+c)(a/b+b/a)>=4c^2+a^2/b+b^2/a$ $a^2/b+b+a+b^2/a+ac/b+bc/a>=4c^2+a^2/b+b^2/a$ $b+a+bc/a+ac/b>=4c^2$ $b>=1>=c>=c^2$ $a>=1>=c>=c^2$ the rest is simple

Solution(I hope this is true): Multiply both sides with ab and we get 3(a^2+b^2)=>4abc^2+a^3+b^3.Because a+b+c=3,the inequality can be rewritten as ba^2+ab^2+c(a^2+b^2)=>4abc^2.Now,observe that a+b=>2c and ab^2+ba^2=ab(a+b)=>2abc and the inequality can be rewritten as (a+b)^2=>4cab which is true because c<=1 and we are done.

$3=a+b+c$ we will replace it in the given inequality. $3*(\frac{a}{b}+\frac{b}{a})=(a+b+c)(\frac{a}{b}+\frac{b}{a})\ge 4*c^2+\frac{a^2}{b}+\frac{b^2}{a}$ Then inequality will be; $(a+b+c)(a^2+b^2) \ge(4*c^2*a*b+a^3+b^3)$ then just simplifize; $a*(b^2)+(a^2)*b+c*(a^2)+c*(b^2) \ge 4c^2*a*b$ <==> $ab*(a+b)+c*(a^2+b^2) \ge ab*(a+b)+ 2*a*b*c \ge 4c^2*a*b$ divide ab; $a+b+2c \ge 4*c^2$ $(1)$ $a \ge c$, $b \ge c$ <==> $a+b \ge 2*c$ from $(1)$; $4*c \ge 4*c^2$ <==> $1 \ge c$ which is true (given condition)

Clearing out the denominators, the inequality becomes \[3(a^2 + b^2) \geq 4abc^2 + a^3 + b^3\] Note that $a^3 + b^3 = (a+b)(a^2 + b^2) - a^2b - ab^2$. Using this, the inequality becomes \[3(a^2 + b^2) \geq 4abc^2 + (a+b)(a^2 + b^2) - a^2b - ab^2\]\[(3 - a - b)(a^2 + b^2) + ab(a + b) \geq 4abc^2\]\[c(a^2 + b^2) + ab(3 - c) \geq 4abc^2\] By AM-GM, we can write $a^2 + b^2 \geq 2ab$. \[c(2ab) + 3ab - abc \geq 4abc^2\]\[2c + 3 - c \geq 4c^2\]\[0 \geq 4c^2 - c - 3 = (4c + 3)(c - 1)\] Because $c$ is in the interval $[0, 1]$, $(4c + 3)(c - 1)$ will be negative, as desired.