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
Snakes
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 $\frac{(x^2+y^2)}{xy}\geq 2$
rmtf1111
19.03.2017 18:02
Let $a=\tan{\alpha}$, $b=\tan{\beta}$ , $c=\tan{\gamma}$ , where $\alpha$ , $\beta$ , $\gamma$ are the measures of the angles of an acute triangle . Let $r$ and $R$ be the radius of incenter and respectively circumcenter of the triangle mentioned.
$$\sum_{\text{cyc}}\sqrt{(1+(\tan{\alpha})^2)(1+(\tan{\beta})^2)} \ge 4+\sqrt{(1+(\tan{\alpha})^2)(1+(\tan{\beta})^2)(1+(\tan{\gamma})^2)} \Longleftrightarrow$$$$\sum_{\text{cyc}}\frac{1}{\cos{\alpha}\cos{\beta}} \ge 4+\frac{1}{\cos{\alpha}\cos{\beta}\cos{\gamma}} \Longleftrightarrow \sum_{\text{cyc}}\cos{\alpha} \ge 1+4\prod \cos{\alpha}
\Longleftrightarrow$$$$1+\frac{r}{R} \ge 1+4\prod \cos{\alpha} \Longleftrightarrow \frac{r}{R} \ge 4\prod \cos{\alpha}$$
$$r=4R\prod \sin{\frac{\alpha}{2}} \implies \frac{r}{R} \ge 4\prod \cos{\alpha} \Longleftrightarrow \prod \sin{\frac{\alpha}{2}} \ge \prod \cos{\alpha} \stackrel{\text{Ravi's substitutions}}{\Longleftrightarrow} $$$$\frac{xyz}{\prod (x+y)} \ge \frac{\prod (x^2+xy+xz-yz)}{\prod (x+y)^2} \Longleftrightarrow xyz\sum_{\text{sym}}x^2y \ge -\sum_{\text{sym}}x^4y^2-2\sum_{\text{cyc}}x^3y^3+2\sum_{\text{cyc}} x^4yz+2\sum_{\text{sym}} x^3y^2z+2x^2y^2z^2 \Longleftrightarrow $$$$2\sum_{\text{cyc}}x^3y^3 + \sum_{\text{sym}} x^4y^2 \ge 2\sum_{\text{cyc}} x^4yz+\sum_{\text{sym}} x^3yz^2$$But from AM-GM we have that $2\sum_{\text{cyc}}x^3y^3 \ge \sum_{\text{sym}} x^3yz^2$ and $\sum_{\text{sym}} x^4y^2 \ge 2\sum_{\text{cyc}} x^4yz$ so we are done.
sqing
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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sqing
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 $$ sin\frac{A}{2}sin\frac{B}{2}sin\frac{C}{2} \ge cosAcosBcosC $$
KantNguyen
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
sqing
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.
sqing
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