Since $\frac{4(ab + bc + ca)^2}{a+b+c} \ge 4(ab + bc + ca)$, it is sufficient to prove $(a + b + c)(ab + bc+ca) + 3abc \ge \frac{4(ab + bc + ca)^2}{a+b+c}$ $\Longleftarrow \sum_{cyc}ab(a-b)^2\ge 0$

parmenides51 wrote: Given positive real numbers $a,b,c$ with $ab+bc+ca\ge a+b+c$ , prove that $$(a + b + c)(ab + bc+ca) + 3abc \ge 4(ab + bc + ca).$$ (I. Gorodnin)

sqing wrote: parmenides51 wrote: Given positive real numbers $a,b,c$ with $ab+bc+ca\ge a+b+c$ , prove that $$(a + b + c)(ab + bc+ca) + 3abc \ge 4(ab + bc + ca).$$ (I. Gorodnin) How do you prove the last one? I think this questions's conclusion is just be done by factorising(except long ones) .

First we will use the inequality that they give us like that; ab+bc+ac>=a+b+c We will write a+b+c instead of ab+ac+bc. We will get (a+b+c)² + 3abc >= 4(ab+bc+ac) Let's open the bracket a²+b²+c²+2ab+2ac+2bc+3abc >= 4ab+4ac+4bc a²+b²+c²+3abc>= 2ab+2ac+2bc a²+b²+c²+3abc-2ab-2ac-2bc >=0 We can see from there (a+b+c)² I wanted to get this so i did ; -(a+b+c)²+3abc+2a²+2b²+2c²>=0 -(a+b+c)²>=-2a²-2b²-2c²-3abc (a+b+c)²<=2a²+2b²+2c²+3abc a²+b²+c²+2ab+2ac+2bc<=2a²+2b²+2c²+3abc 2ab+2ac+2bc<=a²+b²+c²+3abc (1) From Muirhead we can get smth like that; (1,1,0)<=(2,0,0) ab+ac+bc<=a²+b²+c² (2) So we find difference of (2)-(1) ab+ac+bc<=3abc Last thing that we must prove Divide the inequality into abc 1/c+1/b+1/a<=3 for a,b,c>0 only when a=b=c=1 left equality will be equal to 3. And it will be the max value. So 1/a+1/b+1/c <=3 . And we proved