Let $a,b,c$ be positive real numbers that satisfy $a+b+c=abc$. Prove that $$\sqrt{(1+a^2)(1+b^2)}+\sqrt{(1+b^2)(1+c^2)}+\sqrt{(1+a^2)(1+c^2)}-\sqrt{(1+a^2)(1+b^2)(1+c^2)} \ge 4.$$
Problem
Source: Moldova TST 2017, Day 2, Problem 2
Tags: inequalities, algebra
19.03.2017 18:01
My solution: We see that $\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=1$ so when denoting $x=\frac{1}{bc}$; $y=\frac{1}{ca}$; $z=\frac{1}{ab}$ we have $\sum x=1$ and $a^2=\frac{x}{yz}$; $b^2=\frac{y}{xz}$; $c^2=\frac{z}{xy}$, so the inequality becomes: $\sum \sqrt{\left(1+\frac{x}{yz}\right)\left(1+\frac{y}{zx}\right)}\geqslant 4+\sqrt{\prod \left( 1+\frac{x}{yz}\right)}\\ \\ \Longleftrightarrow \\ \\ \sum \frac{1}{z}\cdot \sqrt{z^2+1+\frac{(x^2+y^2)z}{xy}}\geqslant 4+\sqrt{\frac{\prod (x+yz)}{xy\cdot yz\cdot xz}}$ Now, $\prod (x+yz)=x^2y^2z^2+\sum x^2y^2+xyz+xyz(x^2+y^2+z^2)=\\ =(xyz)^2+\sum (xy)^2 +2xyz(1-xy-yz-zx)$, because $(x+y+z)^2=1$, thus: $\prod(x+yz)=(xyz)^2+\sum (xy)^2+2\sum xy\cdot yz-2\sum xyz\cdot xy=\\ =(xy+yz+zx-xyz)^2$. Hence $\sum \frac{1}{x}=1+\frac{\sqrt{\prod (x+yz)}}{xyz}$ and it suffices to prove that: $\sum \frac{1}{z}\cdot \sqrt{z^2+1+\frac{(x^2+y^2)z}{xy}}\geqslant \sum 1+\frac{1}{z}$, which is obviously true as $\frac{(x^2+y^2)}{xy}\geq 2$
19.03.2017 18:02
28.03.2017 07:54
rmtf1111 wrote: Let $a,b,c$ be positive real numbers, that satisfy $a+b+c=abc$. Prove the following inequality: $\sqrt{(1+a^2)(1+b^2)}+\sqrt{(1+b^2)(1+c^2)}+\sqrt{(1+a^2)(1+c^2)}-\sqrt{(1+a^2)(1+b^2)(1+c^2)} \ge 4$ Proof of Zhangyanzong:
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30.03.2017 07:11
rmtf1111 wrote: Let $a,b,c$ be positive real numbers, that satisfy $a+b+c=abc$. Prove the following inequality: $\sqrt{(1+a^2)(1+b^2)}+\sqrt{(1+b^2)(1+c^2)}+\sqrt{(1+a^2)(1+c^2)}-\sqrt{(1+a^2)(1+b^2)(1+c^2)} \ge 4$ The inequality is equivalent to For all acute $ \triangle ABC $ , have $$ sin\frac{A}{2}sin\frac{B}{2}sin\frac{C}{2} \ge cosAcosBcosC $$
30.03.2017 15:56
It's 1 problem proposed in Mathematical Magazine in Viet Nam . Solution: $(1+a^{2})(1+b^{2})(1+c^{2})=(1+a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2})+(a^{2}+b^{2}+c^{2}+a^{2}b^{2}c^{2})=(ab+bc+ca-1)^{2} +(a+b+c-abc)^{2}=(ab+bc+ca)^{2} $ We need proved: $\sum \sqrt{(1+b^{2})(1+c^{2})}\geq ab+bc+ca+3$ The last inequalities correct cause: $\sqrt{(1+a^{2})(1+b^{2})}\geq ab+1$ Q.E.D
30.03.2017 16:05
KantNguyen wrote: It's 1 problem proposed in Mathematical Magazine in Viet Nam . Solution: $(1+a^{2})(1+b^{2})(1+c^{2})=(1+a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2})+(a^{2}+b^{2}+c^{2}+a^{2}b^{2}c^{2})=(ab+bc+ca-1)^{2} +(a+b+c-abc)^{2}=(ab+bc+ca)^{2} $ We need proved: $\sum \sqrt{(1+b^{2})(1+c^{2})}\geq ab+bc+ca+3$ The last inequalities correct cause: $\sqrt{(1+a^{2})(1+b^{2})}\geq ab+1$ Q.E.D Very nice.
30.03.2017 16:21
rmtf1111 wrote: Let $a,b,c$ be positive real numbers, that satisfy $a+b+c=abc$. Prove the following inequality: $\sqrt{(1+a^2)(1+b^2)}+\sqrt{(1+b^2)(1+c^2)}+\sqrt{(1+a^2)(1+c^2)}-\sqrt{(1+a^2)(1+b^2)(1+c^2)} \ge 4$ Mathematical and Youth magazine