RagvaloD wrote: For positive numbers is true that $$ab+ac+bc=a+b+c$$Prove $$a+b+c+1 \geq 4abc$$ https://artofproblemsolving.com/community/c6h547310p3170534 math90 wrote: The inequality is equivalent to $\frac{1}{2a+1}+\frac{1}{2b+1}+\frac{1}{2c+1}\geq 1$ which is true since $\sum_{cyc}\frac{1}{2a+1}=\sum_{cyc}\frac{b^2c^2}{2ab^2c^2+b^2c^2}\geq\frac{(ab+bc+ca)^2}{a^2b^2+b^2c^2+c^2a^2+2abc(ab+bc+ca)}=$ $=\frac{(ab+bc+ca)^2}{a^2b^2+b^2c^2+c^2a^2+2abc(a+b+c)}=1$ bel.jad5 wrote: Obviously $a+b+c\geq 3$ and without loss of generality we can assume that $a \geq b \geq c$ meaning that $a \geq 1$, $a+b>1$ and $c = \frac{a+b-ab}{a+b-1}$ So what we need to prove is: \[ a+b+\frac{a+b-ab}{a+b-1} +1 \geq 4ab\frac{a+b-ab}{a+b-1}\]Simplifying this expression: \[ 4a^2b^2-4ab(a+b)+a^2+b^2+a+b+ab-1 \geq 0\]Or: \[(2ab-a-b)^2 \geq (a-1)(b-1)\]we know that $a\geq 1$ so if $b\leq 1$ then the inequality is obviously true. So let's assume that: $b\geq 1$ then we have: \[ ab+1 \geq a+b\]meaning that: \[ 2ab-a-b \geq a+b-2\]So it is enough to prove that: \[ (a+b-2)^2 \geq (a-1)(b-1)\]which is true because: \[ (a+b-2)^2 = (a-1+b-1)^2 \geq 4(a-1)(b-1) \geq (a-1)(b-1)\]Done.

sqing wrote: RagvaloD wrote: For positive numbers is true that $$ab+ac+bc=a+b+c$$Prove $$a+b+c+1 \geq 4abc$$ https://artofproblemsolving.com/community/c6h547310p3170534 math90 wrote: The inequality is equivalent to

Let $a,b,c$ nonnegative real numbers satisfy $a+b+c=ab+bc+ca.$ Prove that $$\frac{7}{5}\geq \frac{1}{2a+1}+\frac{1}{2b+1}+\frac{1}{2c+1}\geq 1.$$