A certain substitution of $a=\frac{x}{y+z}$, $b=\frac{y}{x+z}$, and $c=\frac{z}{x+y}$ gives you the result. It now suffices to prove that $\sum_{cyc}\sqrt{\frac{x}{y+z}} \ge 2$. WLOG $x+y+z=1$ and use $\sqrt{\frac{x}{1-x}} \ge 2x$.

In turn, we could bound the variables in $[0,4]^3$ and see that by Lagrange Multipliers we are done.

Let $a,b,c$ be positiv real numbers such that $ab+bc+ca+2abc=1.$ Prove that $$\frac 1{8a^2+1}+\frac 1{8b^2+1}+\frac 1{8c^2+1}\ge 1.$$

sqing wrote: Let $a,b,c$ be positiv real numbers such that $ab+bc+ca+2abc=1.$ Prove that $$\frac 1{8a^2+1}+\frac 1{8b^2+1}+\frac 1{8c^2+1}\ge 1.$$ $\iff 4(a^2+b^2+c^2)+1\ge 256^2b^2c^2$ which is obvious.

huricane wrote: Let $a,b,c\ge 0$ be three real numbers such that $ab+bc+ca+2abc=1.$ Prove that $$\sqrt{a}+\sqrt{b}+\sqrt{c}\ge 2$$and determine equality cases. http://www.artofproblemsolving.com/community/c6h1184979p5752974 Let $a,b,c>0$ & $ab+ac+bc+2abc=1$. Prove, that $$\sqrt{ab}+\sqrt{ac}+\sqrt{bc}\leq\frac{3}{2}$$

huricane wrote: Let $a,b,c\ge 0$ be three real numbers such that $$ab+bc+ca+2abc=1.$$Prove that $\sqrt{a}+\sqrt{b}+\sqrt{c}\ge 2$ and determine equality cases. By lagrange theorem we easily get that $\sqrt{a}+\sqrt{b}+\sqrt{c}\ge \frac{3\sqrt{2}}{2}$ and the equality occurs when $a=b=c=\frac{1}{2}$

K.N wrote: huricane wrote: Let $a,b,c\ge 0$ be three real numbers such that $$ab+bc+ca+2abc=1.$$Prove that $\sqrt{a}+\sqrt{b}+\sqrt{c}\ge 2$ and determine equality cases. By lagrange theorem we easily get that $\sqrt{a}+\sqrt{b}+\sqrt{c}\ge \frac{3\sqrt{2}}{2}$ and the equality occurs when $a=b=c=\frac{1}{2}$ how about $a=0,b=c=1$?

Reynan wrote: K.N wrote: huricane wrote: Let $a,b,c\ge 0$ be three real numbers such that $$ab+bc+ca+2abc=1.$$Prove that $\sqrt{a}+\sqrt{b}+\sqrt{c}\ge 2$ and determine equality cases. By lagrange theorem we easily get that $\sqrt{a}+\sqrt{b}+\sqrt{c}\ge \frac{3\sqrt{2}}{2}$ and the equality occurs when $a=b=c=\frac{1}{2}$ how about $a=0,b=c=1$? This works when $a,b,c>0$ those cases must be checked separate

K.N wrote: huricane wrote: Let $a,b,c\ge 0$ be three real numbers such that $$ab+bc+ca+2abc=1.$$Prove that $\sqrt{a}+\sqrt{b}+\sqrt{c}\ge 2$ and determine equality cases. By lagrange theorem we easily get that $\sqrt{a}+\sqrt{b}+\sqrt{c}\ge \frac{3\sqrt{2}}{2}$ and the equality occurs when $a=b=c=\frac{1}{2}$ Try $c=0.00000000000000001, \ \ a=b$

huricane wrote: Let $a,b,c\ge 0$ be three real numbers such that $$ab+bc+ca+2abc=1.$$Prove that $\sqrt{a}+\sqrt{b}+\sqrt{c}\ge 2$ and determine equality cases. $\left[ {Vasile\,\,Cirtoaje} \right]$: Let a, b, c non-negative, prove that: $\sqrt {\frac{a}{{b + c}}} + \sqrt {\frac{b}{{c + a}}} + \sqrt {\frac{c}{{a + b}}} \ge 2$.

Nguyenngoctu wrote: $\left[ {Vasile\,\,Cirtoaje} \right]$: Let a, b, c non-negative, prove that: $\sqrt {\frac{a}{{b + c}}} + \sqrt {\frac{b}{{c + a}}} + \sqrt {\frac{c}{{a + b}}} \ge 2$. I think this inequality was before than Vasile Cirtoaje born.

Let $a=\frac{x}{y+z}$, $b=\frac{y}{x+z}$, $c=\frac{z}{x+y}$, we will prove that: $\sum \sqrt{\frac{x}{y+z}} \ge 2$ If $x=0$ or $y=0$, $z=0$ it 's very easy to prove by AM-GM. $\sum \sqrt{\frac{x}{y+z}}= \sum \frac{x}{\sqrt{x(y+z)}}\ge \sum \frac{2x}{x+y+z}=2$ ( By AM-GM: $2\sqrt{x(y+z)} \le x+y+z$)

K.N wrote: huricane wrote: Let $a,b,c\ge 0$ be three real numbers such that $$ab+bc+ca+2abc=1.$$Prove that $\sqrt{a}+\sqrt{b}+\sqrt{c}\ge 2$ and determine equality cases. By lagrange theorem we easily get that $\sqrt{a}+\sqrt{b}+\sqrt{c}\ge \frac{3\sqrt{2}}{2}$ and the equality occurs when $a=b=c=\frac{1}{2}$ I can't get why my solution is not correct Can some one explain for me?! What should we do if we want to solve it by Lagrange method?!

K.N wrote: I can't get why my solution is not correct Can some one explain for me?! Try $a=b=\frac{15}{16}, \ \ c=\frac{1}{30}$. Are you still going to prove wrong inequality ?

arqady wrote: Nguyenngoctu wrote: $\left[ {Vasile\,\,Cirtoaje} \right]$: Let a, b, c non-negative, prove that: $\sqrt {\frac{a}{{b + c}}} + \sqrt {\frac{b}{{c + a}}} + \sqrt {\frac{c}{{a + b}}} \ge 2$. I think this inequality was before than Vasile Cirtoaje born. We have $a\not=0,b\not=0,c\not=0$. We'll show that $\sqrt{\frac{a}{b+c}}\ge \frac{2a}{a+b+c}$,for every $a,b,c$ nonegative numbers. We have $\sqrt{\frac{a}{b+c}}\ge \frac{2a}{a+b+c} \Longleftrightarrow \frac{a}{b+c}\ge (\frac{2a}{a+b+c})^2 \Longleftrightarrow (a+b+c)^2\ge 4a(b+c) \Longleftrightarrow (b+c-a)^2\ge 0$ with equality iff $a=b+c$. Now we easily obtain $\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{a+c}}+\sqrt{\frac{c}{a+b}}\ge \frac{2(a+b+c)}{a+b+c}=2$ with equality if and only if $a=b+c,b=a+c,c=a+b$ $\Longrightarrow$ $a+b+c=0$ $\Longrightarrow$ $a=b=c=0$ $which$ $is$ $impossible.$ $\Longrightarrow$.$\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{a+c}}+\sqrt{\frac{c}{a+b}}$>$2$

mudok wrote: K.N wrote: I can't get why my solution is not correct Can some one explain for me?! Try $a=b=\frac{15}{16}, \ \ c=\frac{1}{30}$. Are you still going to prove wrong inequality ? I know what you say But i mean if you have a correct solution by Lagrange method i'll be satisfied if you tell me Thanks:)

Orkhan-Ashraf_2002 wrote: arqady wrote: Nguyenngoctu wrote: $\left[ {Vasile\,\,Cirtoaje} \right]$: Let a, b, c non-negative, prove that: $\sqrt {\frac{a}{{b + c}}} + \sqrt {\frac{b}{{c + a}}} + \sqrt {\frac{c}{{a + b}}} \ge 2$. I think this inequality was before than Vasile Cirtoaje born. We'll show that $\sqrt{\frac{a}{b+c}}\ge \frac{2a}{a+b+c}$,for every $a,b,c$ nonegative numbers. We have $\sqrt{\frac{a}{b+c}}\ge \frac{2a}{a+b+c} \Longleftrightarrow \frac{a}{b+c}\ge (\frac{2a}{a+b+c})^2 \Longleftrightarrow (a+b+c)^2\ge 4a(b+c) \Longleftrightarrow (b+c-a)^2\ge 0$ with equality iff $a=b+c$. Now we easily obtain $\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{a+c}}+\sqrt{\frac{c}{a+b}}\ge \frac{2(a+b+c)}{a+b+c}=2$ with equality if and only if $a=b+c,b=a+c,c=a+b$ $\Longrightarrow$ $a+b+c=0$ $\Longrightarrow$ $a=b=c=0$. $a=b=c=0$ then $0>2$

thinhrost1 wrote: Let $a=\frac{x}{y+z}$, $b=\frac{y}{x+z}$, $c=\frac{z}{x+y}$, we will prove that: $\sum \sqrt{\frac{x}{y+z}} \ge 2$ If $x=0$ or $y=0$, $z=0$ it 's very easy to prove by AM-GM. $\sum \sqrt{\frac{x}{y+z}}= \sum \frac{x}{\sqrt{x(y+z)}}\ge \sum \frac{2x}{x+y+z}=2$ ( By AM-GM: $2\sqrt{x(y+z)} \le x+y+z$) EQUALITY?

Nguyenngoctu wrote: huricane wrote: Let $a,b,c\ge 0$ be three real numbers such that $$ab+bc+ca+2abc=1.$$Prove that $\sqrt{a}+\sqrt{b}+\sqrt{c}\ge 2$ and determine equality cases. $\left[ {Vasile\,\,Cirtoaje} \right]$: Let a, b, c non-negative, prove that: $\sqrt {\frac{a}{{b + c}}} + \sqrt {\frac{b}{{c + a}}} + \sqrt {\frac{c}{{a + b}}} \ge 2$. This inequality wrong.Right inequality $\Longrightarrow$ $\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{a+c}}+\sqrt{\frac{c}{a+b}}>2$

Orkhan-Ashraf_2002 wrote: This inequality wrong.Right inequality $\Longrightarrow$ $\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{a+c}}+\sqrt{\frac{c}{a+b}}>2$ Hi Orxan,then you proof is right? I think your proof is false.

Orkhan-Ashraf_2002 wrote: thinhrost1 wrote: Let $a=\frac{x}{y+z}$, $b=\frac{y}{x+z}$, $c=\frac{z}{x+y}$, we will prove that: $\sum \sqrt{\frac{x}{y+z}} \ge 2$ If $x=0$ or $y=0$, $z=0$ it 's very easy to prove by AM-GM. $\sum \sqrt{\frac{x}{y+z}}= \sum \frac{x}{\sqrt{x(y+z)}}\ge \sum \frac{2x}{x+y+z}=2$ ( By AM-GM: $2\sqrt{x(y+z)} \le x+y+z$) EQUALITY? In the first case: equality when $x=0, y=z$

thinhrost1 wrote: Orkhan-Ashraf_2002 wrote: thinhrost1 wrote: Let $a=\frac{x}{y+z}$, $b=\frac{y}{x+z}$, $c=\frac{z}{x+y}$, we will prove that: $\sum \sqrt{\frac{x}{y+z}} \ge 2$ If $x=0$ or $y=0$, $z=0$ it 's very easy to prove by AM-GM. $\sum \sqrt{\frac{x}{y+z}}= \sum \frac{x}{\sqrt{x(y+z)}}\ge \sum \frac{2x}{x+y+z}=2$ ( By AM-GM: $2\sqrt{x(y+z)} \le x+y+z$) EQUALITY? In the first case: equality when $x=0, y=z$ Look my post above.I wrote equality if and only if $a=b+c,b=a+c,c=a+b$ $\Longrightarrow$ $a=b=c=0$.$Which$ $is$ $impossible$.Then equality is not a state;Because $a\not=0$;$b\not=0$;$c\not=0$.

Orkhan-Ashraf_2002 wrote: Look my post above.I wrote equality if and only if $a=b+c,b=a+c,c=a+b$ $\Longrightarrow$ $a=b=c=0$.$Which$ $is$ $impossible$.Then equality is not a state; Orxan,you are false.The equality of $(x,y,z)=(0,y,y),(0,z,z),(x,0,x),(z,0,z),(x,x,0),(y,y,0)$

Ferid.---. wrote: Orkhan-Ashraf_2002 wrote: Look my post above.I wrote equality if and only if $a=b+c,b=a+c,c=a+b$ $\Longrightarrow$ $a=b=c=0$.$Which$ $is$ $impossible$.Then equality is not a state; Orxan,you are false.The equality of $(x,y,z)=(0,y,y),(0,z,z),(x,0,x),(z,0,z),(x,x,0),(y,y,0)$ $x\not=0;y\not=0;z\not=0.$If$ x=0$ $\Longrightarrow a=0,$if$ y=0 \Longrightarrow b=0,$if$ z=0 \Longrightarrow c=0$.It is a imbalance.

Orkhan-Ashraf_2002 wrote: arqady wrote: Nguyenngoctu wrote: $\left[ {Vasile\,\,Cirtoaje} \right]$: Let a, b, c non-negative, prove that: $\sqrt {\frac{a}{{b + c}}} + \sqrt {\frac{b}{{c + a}}} + \sqrt {\frac{c}{{a + b}}} \ge 2$. I think this inequality was before than Vasile Cirtoaje born. We have $a\not=0,b\not=0,c\not=0$. We'll show that $\sqrt{\frac{a}{b+c}}\ge \frac{2a}{a+b+c}$,for every $a,b,c$ nonegative numbers. We have $\sqrt{\frac{a}{b+c}}\ge \frac{2a}{a+b+c} \Longleftrightarrow \frac{a}{b+c}\ge (\frac{2a}{a+b+c})^2 \Longleftrightarrow (a+b+c)^2\ge 4a(b+c) \Longleftrightarrow (b+c-a)^2\ge 0$ with equality iff $a=b+c$. Now we easily obtain $\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{a+c}}+\sqrt{\frac{c}{a+b}}\ge \frac{2(a+b+c)}{a+b+c}=2$ with equality if and only if $a=b+c,b=a+c,c=a+b$ $\Longrightarrow$ $a+b+c=0$ $\Longrightarrow$ $a=b=c=0$ $which$ $is$ $impossible.$ $\Longrightarrow$.$\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{a+c}}+\sqrt{\frac{c}{a+b}}$>$2$ If $a+b+c=0$ then we can't have $\frac{2a}{a+b+c}$, so we have to seperate into two cases: + all of $a,b,c \ne 0$ +or $a \ne 0$, or $b\ne 0$, or $c\ne 0$ Cearly in the first case there aren't any a,b,c hold equality The second equality when $a=0$, $b=c$ ...

Dear thinhrost. You wrote $a\not=0;b\not=0;c\not=0$, after you wrote equality when $a=0,b=c.$

Orxan your solution is wrong,because is impossible.Right solution is #thinhrost1.

Ferid.---. wrote: Orxan your solution is wrong,because is impossible.Right solution is #thinhrost1. Prove that Farid my solution is wrong.

Orkhan-Ashraf_2002 wrote: Ferid.---. wrote: Orxan your solution is wrong,because is impossible.Right solution is #thinhrost1. Prove that Farid my solution is wrong. It's easy to prove Orkhan-Ashraf_2002 wrote: arqady wrote: Nguyenngoctu wrote: $\left[ {Vasile\,\,Cirtoaje} \right]$: Let a, b, c non-negative, prove that: $\sqrt {\frac{a}{{b + c}}} + \sqrt {\frac{b}{{c + a}}} + \sqrt {\frac{c}{{a + b}}} \ge 2$. I think this inequality was before than Vasile Cirtoaje born. We have $a\not=0,b\not=0,c\not=0$. We'll show that $\sqrt{\frac{a}{b+c}}\ge \frac{2a}{a+b+c}$,for every $a,b,c$ nonegative numbers. We have $\sqrt{\frac{a}{b+c}}\ge \frac{2a}{a+b+c} \Longleftrightarrow \frac{a}{b+c}\ge (\frac{2a}{a+b+c})^2 \Longleftrightarrow (a+b+c)^2\ge 4a(b+c) \Longleftrightarrow (b+c-a)^2\ge 0$ with equality iff $a=b+c$. Now we easily obtain $\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{a+c}}+\sqrt{\frac{c}{a+b}}\ge \frac{2(a+b+c)}{a+b+c}=2$ with equality if and only if $a=b+c,b=a+c,c=a+b$ $\Longrightarrow$ $a+b+c=0$ $\Longrightarrow$ $a=b=c=0$ $which$ $is$ $impossible.$ $\Longrightarrow$.$\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{a+c}}+\sqrt{\frac{c}{a+b}}$>$2$ At here you said: $\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{a+c}}+\sqrt{\frac{c}{a+b}}$>$2$ for all non-negatives $a,b,c$( it means: $a,b,c \ge 0$) But when a=0,$ b=c \ne0$ then $\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{a+c}}+\sqrt{\frac{c}{a+b}}=2$ Contradiction

Yes, i solved this question when $a,b,c>0$. $\Longrightarrow \sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{a+c}}$+$\sqrt{\frac{c}{a+b}}$>2.Sorry,i don't consider $a,b,c\ge 0.$

Let $a,b,c$ be positiv real numbers such that $ab+bc+ca+2abc=1.$ Prove that $$ 4a+b+c\ge 2.$$(Mathematical Olympiad 2015 A.Petroupolis Hg (9th grade))

sqing wrote: Let $a,b,c$ be positiv real numbers such that $ab+bc+ca+2abc=1.$ Prove that $$ 4a+b+c\ge 2.$$ Let $a=\frac{x}{y+z}$ and $b=\frac{y}{x+z}$, where $x$, $y$ and $z$ are positives. Hence, $c=\frac{z}{x+y}$ and by C-S we obtain: $4a+b+c-2=\frac{4x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}-2\geq\frac{(2x+y+z)^2}{2(xy+xz+yz)}-2=\frac{4x^2+(y-z)^2}{2(xy+xz+yz)}\geq0$.

anantmudgal09 wrote: In turn, we could bound the variables in $[0,4]^3$ and see that by Lagrange Multipliers we are done. What are Langrange Multipliers ?

kk108 wrote: What are Langrange Multipliers ? See here: https://en.wikipedia.org/wiki/Lagrange_multiplier

arqady wrote: sqing wrote: Let $a,b,c$ be positiv real numbers such that $ab+bc+ca+2abc=1.$ Prove that $$ 4a+b+c\ge 2.$$ Let $a=\frac{x}{y+z}$ and $b=\frac{y}{x+z}$, where $x$, $y$ and $z$ are positives. Hence, $c=\frac{z}{x+y}$ and by C-S we obtain: $4a+b+c-2=\frac{4x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}-2\geq\frac{(2x+y+z)^2}{2(xy+xz+yz)}-2=\frac{4x^2+(y-z)^2}{2(xy+xz+yz)}\geq0$. Very nice. Proof of matha: The inequality is strict unless the pronunciation saying "non-negative". There are positive $\displaystyle {a, b, c}$ that $\displaystyle {x = \frac {a} {b + c}, y = \frac {b} {c + a}, z = \frac {c} {a + b},}$ when the verifiable written $\displaystyle {\frac {4a} {b + c} + \frac {b} {c + a} + \frac {c} {a + b} \geq 2.}$ From Cauchy-Schwarz is $\displaystyle{\frac{4a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=\frac{4a^2}{ab+ac}+\frac{b^2}{bc+ab}+\frac{c^2}{ac+bc}\geq \frac {(2a + b + c) ^ 2} {2 (ab + bc + ca)}}$ it is sufficient to prove that $\displaystyle {(2a + b + c) ^ 2 \geq 4 (ab + bc + ca)}$ which is equivalent to the apparent$ \displaystyle {4a ^ 2 + (b-c) ^ 2 \geq 0.}$

sqing wrote: Let $a,b,c$ be positiv real numbers such that $ab+bc+ca+2abc=1.$ Prove that $$ 4a+b+c\ge 2.$$(Mathematical Olympiad 2015 A.Petroupolis Hg (9th grade))

Let $a,b,c$ be positiv real numbers such that $ab+bc+ca+2abc=1.$ Prove that $$ a^2+b+c\ge \frac{5}{4}.$$(By Grotex)

Grotex wrote: Let $a,b,c$ be positiv real numbers such that $ab+bc+ca+2abc=1.$ Prove that $$ a^2+b+c\ge \frac{5}{4}.$$(By Grotex) Very nice.

sqing wrote: Let $a,b,c$ be positiv real numbers such that $ab+bc+ca+2abc=1.$ Prove that $$\frac 1{8a^2+1}+\frac 1{8b^2+1}+\frac 1{8c^2+1}\ge 1.$$ Let $t = \sqrt[3]{{abc}}$，Since $1 = ab + bc + ca + 2abc \ge 3\sqrt[3]{{\left( {abc} \right)^2 }} + 2abc = 3t^2 + 2t^3 \Rightarrow 2t^3 + 3t^2 - 1 \le 0 \Rightarrow $ $\left( {2t - 1} \right)\left( {t + 1} \right)^2 \le 0 \Leftrightarrow t \le \frac{1}{2} \Leftrightarrow abc \le \frac{1}{8}$， Since $abc \le \frac{1}{8} \Rightarrow \exists k,x,y,z > 0:\left( {a^2 ,b^2 ,c^2 } \right) = \left( {\frac{{kyz}}{{x^2 }},\frac{{kzx}}{{y^2 }},,\frac{{kxy}}{{z^2 }}} \right)$,and $0 < k \le \frac{1}{4}$， By C-S We have $\sum {\frac{1}{{8a^2 + 1}}} = \sum {\frac{{x^2 }}{{x^2 + 8kyz}}} = \ge \frac{{\left( {\sum x } \right)^2 }}{{\sum {x^2 + 24kxyz} }} \ge \frac{{\sum {x^2 + 6xyz} }}{{\sum {x^2 + 24 \times \frac{1}{4}xyz} }} = 1$ Generalization Let $a,b,c,\lambda ,\mu $ are positive real numbers such that $ab + bc + ca + 2abc = 1$ and $\lambda \le 2^{\mu + 1} $ .Prove that $$\frac{1}{{\lambda a^\mu + 1}} + \frac{1}{{\lambda b^\mu + 1}} + \frac{1}{{\lambda c^\mu + 1}} \ge 1{\rm{ }}$$

sqing wrote: Let $a,b,c$ be positiv real numbers such that $ab+bc+ca+2abc=1.$ Prove that $$ 4a+b+c\ge 2.$$(Mathematical Olympiad 2015 A.Petroupolis Hg (9th grade)) Let $a,b,c$ be positiv real numbers such that $ab+bc+ca+2abc=1.$ For any positive numbers $x,y,z,$ prove that $$ xa+yb+zc\ge \frac 12(\sqrt{x}+\sqrt{y}+\sqrt{z})^2-(x+y+z).$$

huricane wrote: Let $a,b,c\ge 0$ be three real numbers such that $$ab+bc+ca+2abc=1.$$Prove that $\sqrt{a}+\sqrt{b}+\sqrt{c}\ge 2$ and determine equality cases. Let $ x,y,z $ be three positive real numbers such that $ x^2+y^2+z^2+3=2(xy+yz+zx) . $ Show that $$ \sqrt{xy}+\sqrt{yz}+\sqrt{zx}\ge 3, $$Stars of Mathematics 2017, Juniors

huricane wrote: Let $a,b,c\ge 0$ be three real numbers such that $$ab+bc+ca+2abc=1.$$Prove that $$\sqrt{a}+\sqrt{b}+\sqrt{c}\ge 2$$and determine equality cases. Let $a,b,c \ge 0 : ab+bc+ca=1$. Prove that: $$\sqrt{a}+\sqrt{b}+\sqrt{c} \ge 2$$