Define $a=\frac{x}{y}$ $b=\frac{y}{z}$. After opening the brackets, the statement becomes $2\sqrt{ab+a+b+\frac{1}{a}+\frac{1}{b}+\frac{1}{ab}+3}$$-$$\sqrt{ab+a+b+1}$, which can be groupped as $2\sqrt{(ab+a+b+1)+\frac{ab+a+b+1}{ab}+1}$$-$$\sqrt{ab+a+b+1}$ Now, substituting $ab+a+b+1=4m^2$ $(1)$ and $n^2-4m^2-1=\frac{ab+a+b+1}{ab}$ $(2)$, the main part obviously becomes $2(n-m)$. So, it's sufficient to find the minimum value of $2(n-m)$ Dividing $(1)$ to $(2)$, we have $ab=\frac{4m^2}{n^2-4m^2-1}$ $(3)$ Using Cauchy-Schwarz Inequality, we have $(a+1).(b+1)\geq(\sqrt{ab}+1)^2$, which turns into $\sqrt{ab+a+b+1}\geq\sqrt{ab}+1$ . Using $(1)$ and $(3)$, we get $2m\geq\sqrt{\frac{4m^2}{n^2-4m^2-1}}+1$ $\sqrt{n^2-4m^2-1}\geq\frac{2m}{2m-1}=1+\frac{1}{2m-1}$ $n^2-4m^2-1\geq\left(\frac{1}{2m-1}\right)^2$ $n^2\geq4m^2+1+\left(\frac{1}{2m-1}\right)^2=(4m^2-4m+1)+2.(2m-1).(\frac{2m}{2m-1})+\left(\frac{1}{2m-1}\right)^2$ $=((2m-1)+(1+\frac{1}{2m-1}))^2$ $=(2m+\frac{1}{2m-1})^2$ After taking the square roots and substracting $m$, we get $n-m\geq m+\frac{1}{2m-1}$. Multiplying with 2, we have $2(n-m)\geq 2m+\frac{2}{2m-1}$ $=$ $(2m-1)+\frac{2}{2m-1}+1\geq2\sqrt{2}+1$ (Am-Gm) But the part $2(n-m)$ is indeed our original statement. So the conclusion follows (Equality occurs iff $a=b=\sqrt{2}$, which means $(x,y,z)=(2z,z\sqrt{2},z)$

Pqrq wrote: Define $a=\frac{x}{y}$ $b=\frac{y}{z}$. After opening the brackets, the statement becomes $2\sqrt{ab+a+b+\frac{1}{a}+\frac{1}{b}+\frac{1}{ab}+3}$$-$$\sqrt{ab+a+b+1}$ Now, let $ab=const.$ Thus, our expression increases as a function of $a+b$, which says that it's enough to assume $a=b$.

arqady wrote: Pqrq wrote: Define $a=\frac{x}{y}$ $b=\frac{y}{z}$. After opening the brackets, the statement becomes $2\sqrt{ab+a+b+\frac{1}{a}+\frac{1}{b}+\frac{1}{ab}+3}$$-$$\sqrt{ab+a+b+1}$ Now, let $ab=const.$ Thus, our expression increases as a function of $a+b$, which says that it's enough to assume $a=b$. I wrote this in the real exam and I wonder how many points I can take sir.

electrovector, for $b=a$ we obtain: $$2\sqrt{ab+a+b+\frac{1}{a}+\frac{1}{b}+\frac{1}{ab}+3}-\sqrt{ab+a+b+1}=$$$$=\frac{2(a^2+a+1)}{a}-a-1=1+a+\frac{2}{a}\geq1+2\sqrt2.$$

arqady wrote: Pqrq wrote: Define $a=\frac{x}{y}$ $b=\frac{y}{z}$. After opening the brackets, the statement becomes $2\sqrt{ab+a+b+\frac{1}{a}+\frac{1}{b}+\frac{1}{ab}+3}$$-$$\sqrt{ab+a+b+1}$ Now, let $ab=const.$ Thus, our expression increases as a function of $a+b$, which says that it's enough to assume $a=b$.

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TerenceTao11235, let $a+b=2u$ and $ab=v^2$. Thus, by AM-GM $u^2\geq v^2$ and $$2\sqrt{ab+a+b+\frac{1}{a}+\frac{1}{b}+\frac{1}{ab}+3}-\sqrt{(a+1)(b+1)}=$$$$=2\sqrt{2u+\frac{2u}{v^2}+v^2+\frac{1}{v^2}+3}-\sqrt{v^2+2u+1}=f(u).$$Now, $$f'(u)=\frac{2+\frac{2}{v^2}}{\sqrt{2u+\frac{2u}{v^2}+v^2+\frac{1}{v^2}+3}}-\frac{1}{\sqrt{v^2+2u+1}}>0.$$Id est, it's enough to understand, what happens for $u^2=v^2$ or for $a=b$.

arqady wrote: electrovector, for $b=a$ we obtain: $$2\sqrt{ab+a+b+\frac{1}{a}+\frac{1}{b}+\frac{1}{ab}+3}-\sqrt{ab+a+b+1}=$$$$=\frac{2(a^2+a+1)}{a}-a-1=1+a+\frac{2}{a}\geq1+2\sqrt2.$$ I did the same thing in the exam. All the solution is same however I feel like I didn't express myself clearly in the exam because of saying "take a+b=m and ab=n, write the expression in terms of m and let this be f(m). We can easily see f(m+d)>f(m)."

arqady wrote: TerenceTao11235, let $a+b=2u$ and $ab=v^2$. Thus, by AM-GM $u^2\geq v^2$ and $$2\sqrt{ab+a+b+\frac{1}{a}+\frac{1}{b}+\frac{1}{ab}+3}-\sqrt{(a+1)(b+1)}=$$$$=2\sqrt{2u+\frac{2u}{v^2}+v^2+\frac{1}{v^2}+3}-\sqrt{v^2+2u+1}=f(u).$$Now, $$f'(u)=\frac{2+\frac{2}{v^2}}{\sqrt{2u+\frac{2u}{v^2}+v^2+\frac{1}{v^2}+3}}-\frac{1}{\sqrt{v^2+2u+1}}>0.$$Id est, it's enough to understand, what happens for $u^2=v^2$ or for $a=b$.

just bash for $u>v$?

Yes. We can assume that $v>0$, which gives $u\geq v$.

electrovector wrote: If $x, y, z$ are positive real numbers find the minimum value of $$2\sqrt{(x+y+z) \left( \frac{1}{x}+ \frac{1}{y} + \frac{1}{z} \right)} - \sqrt{ \left( 1+ \frac{x}{y} \right) \left( 1+ \frac{y}{z} \right)}$$ $$LHS=2\sqrt{\left(x+y+z \right)\left(\frac{1}{y}+\frac{1}{z}+\frac{1}{x} \right)}-\sqrt{\left(1+\frac{x}{y} \right)\left(1+\frac{y}{z} \right)}$$$$LHS \ge 2\left(\sqrt{\left(x+y \right)\left(\frac{1}{y}+\frac{1}{z} \right)}+\sqrt{\frac{z}{x}} \right)-\sqrt{\left(1+\frac{x}{y} \right)\left(1+\frac{y}{z} \right)}=\sqrt{\left(1+\frac{x}{y} \right)\left(1+\frac{y}{z} \right)}+2\sqrt{\frac{z}{x}}$$$$LHS \ge 1+\sqrt{\frac{x}{z}}+2\sqrt{\frac{z}{x}} \ge 1+2\sqrt{2}.$$

Let $a,b,c$ are positive real numbers. Prove that $$2\sqrt{(a+b+c) \left( \frac{1}{a}+ \frac{1}{b} + \frac{1}{c} \right)} - \sqrt{ \left( 1+ \frac{a}{b} \right) \left( 1+ \frac{b}{c} \right)} \geq2\sqrt 2+1.$$Equality holds when $a:b:c=2:\sqrt 2:1.$

anhduy98 wrote: electrovector wrote: If $x, y, z$ are positive real numbers find the minimum value of $$2\sqrt{(x+y+z) \left( \frac{1}{x}+ \frac{1}{y} + \frac{1}{z} \right)} - \sqrt{ \left( 1+ \frac{x}{y} \right) \left( 1+ \frac{y}{z} \right)}$$ $$LHS=2\sqrt{\left(x+y+z \right)\left(\frac{1}{y}+\frac{1}{z}+\frac{1}{x} \right)}-\sqrt{\left(1+\frac{x}{y} \right)\left(1+\frac{y}{z} \right)}$$$$LHS \ge 2\left(\sqrt{\left(x+y \right)\left(\frac{1}{y}+\frac{1}{z} \right)}+\sqrt{\frac{z}{x}} \right)-\sqrt{\left(1+\frac{x}{y} \right)\left(1+\frac{y}{z} \right)}=\sqrt{\left(1+\frac{x}{y} \right)\left(1+\frac{y}{z} \right)}+2\sqrt{\frac{z}{x}}$$$$LHS \ge 1+\sqrt{\frac{x}{z}}+2\sqrt{\frac{z}{x}} \ge 1+2\sqrt{2}.$$ Pretty clean, thanks for the solution!

electrovector wrote: If $x, y, z$ are positive real numbers find the minimum value of $$2\sqrt{(x+y+z) \left( \frac{1}{x}+ \frac{1}{y} + \frac{1}{z} \right)} - \sqrt{ \left( 1+ \frac{x}{y} \right) \left( 1+ \frac{y}{z} \right)}$$ Solution of Zhangyanzong: $$2\sqrt{(x+y+z) \left( \frac{1}{x}+ \frac{1}{y} + \frac{1}{z} \right)} - \sqrt{ \left( 1+ \frac{x}{y} \right) \left( 1+ \frac{y}{z} \right)}$$$$=\sqrt{\left( (x+y)+(x+y+2z) \right) \left(\left (\frac{1}{y}+ \frac{1}{z}\right) +\left (\frac{2}{x}+\frac{1}{y}+ \frac{1}{z}\right) \right)} - \sqrt{ \left( 1+ \frac{x}{y} \right) \left( 1+ \frac{y}{z} \right)}$$$$\geq \sqrt{\left( x+2z+y\right) \left(\frac{2}{x}+ \frac{1}{z}+\frac{1}{y}\right)} \geq2\sqrt 2+1$$

Let $a, b,c,d$ are positive real numbers. Prove that $$3\sqrt{(a+b+c+d)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\right)}-\sqrt{\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)}-\sqrt{\left(1+\frac{c}{d}\right)\left(1+\frac{d}{a}\right)}\geq 8$$(Lijvzhi)

Let $a,b,c$ are positive real numbers. Prove that

Let $a,b,c$ are positive real numbers. Prove that $$ (a+b+c)(\frac{1}{a}+\frac{1}{b})(1+\frac{a}{b})(1+\frac{b}{c})(1+\frac{c}{a}) \geq 22+10\sqrt 5$$$$ (a+b+c)(\frac{1}{a}+\frac{1}{b} +\frac{1}{c})(1+\frac{a}{b})(1+\frac{b}{c})>27$$$$ (\frac{b}{a}+\frac{c}{b} +\frac{a}{c})(1+\frac{a}{b})(1+\frac{b}{c}) >9$$

sqing wrote: Let $a, b,c,d$ are positive real numbers. Prove that $$3\sqrt{(a+b+c+d)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\right)}-\sqrt{\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)}-\sqrt{\left(1+\frac{c}{d}\right)\left(1+\frac{d}{a}\right)}\geq 8$$(Lijvzhi) Same method. $$\sqrt{(a+b+c+d)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\right)} \geq \sqrt{(a+b) \left( \frac{1}{b} + \frac{1}{c} \right)}+ \sqrt{(c+d) \left( \frac{1}{a}+\frac{1}{d} \right)}$$Thus we are left with $2\sqrt{(a+b+c+d)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\right)}$ which has a minimum value of $8$.

sqing wrote: Let $a,b,c$ are positive real numbers. Prove that $$3\sqrt{(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})}-\sqrt{(1+\frac{a}{b})(1+\frac{b}{c})}\geq 2(\sqrt 6+1)$$$$3\sqrt{(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})}-2\sqrt{(1+\frac{a}{b})(1+\frac{b}{c})}\geq 2\sqrt 3+1$$$$4\sqrt{(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})}-3\sqrt{(1+\frac{a}{b})(1+\frac{b}{c})}\geq 5$$$$\sqrt{(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})}+\sqrt{(1+\frac{a}{b})(1+\frac{b}{c})}\geq 2(\sqrt 2+1)$$$$\sqrt{(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})}+3\sqrt{(1+\frac{a}{b})(1+\frac{b}{c})}\geq 8$$ All can be solved by the same method.

sqing wrote: Could you please write one of the proofs in detail? Thanks. [/hide] Of course. Let's solve the first ineq as I numbered above. There are two main points to see while solving this problem. First one is trying to use Cauchy-Scwarz in an appropriate form to make inequality simpler. This part is obvious because of the fact that we have squareroot of product of sums. The annoying second part in this inequality is the part $\sqrt{\left( 1+ \frac{a}{b} \right) \left( 1+ \frac{b}{c} \right)}$ since it has a minus sign in front of it. Hence we must get rid of this. To make this annoying part look like the normal part we arrange it a little bit and it looks like this $$\sqrt{\left( 1+ \frac{a}{b} \right) \left( 1+ \frac{b}{c} \right)}=\sqrt{(a+b) \left( \frac{1}{b} +\frac{1}{c} \right)}$$That is where the magic happens. We see this is similar to $\sqrt{(a+b+c) \left( \frac{1}{a}+\frac{1}{b}+ \frac{1}{c} \right)}$ and applying C-S to this we will get $$3 \sqrt{(a+b+c) \left( \frac{1}{a}+\frac{1}{b}+ \frac{1}{c} \right)} \geq 3\sqrt{(a+b) \left(\frac{1}{b}+\frac{1}{c} \right)} +3\sqrt{\frac{c}{a} } \Rightarrow 3 \sqrt{(a+b+c) \left( \frac{1}{a}+\frac{1}{b}+ \frac{1}{c} \right)} -\sqrt{\left( 1+\frac{a}{b} \right) \left(1+ \frac{b}{c} \right)} \geq 2\sqrt{(a+b) \left(\frac{1}{b}+\frac{1}{c} \right)} +3\sqrt{\frac{c}{a}}$$With the same motivation we have to get something simpler. This time it is easier to see though. We will cancel $b$ using C-S and use an easy AM-GM which will finish the problem at the end. $$2\sqrt{(b+a) \left(\frac{1}{b}+\frac{1}{c} \right) }+3\sqrt{\frac{c}{a}} \geq 2+ 2\sqrt{\frac{a}{c}}+3\sqrt{\frac{c}{a}} \geq 2+2 \sqrt{6}$$and we are done.

Generalization 1 Let $x_{1},x_2,\cdots,x_{2k+1}$ be positive reals ($k\geq 1$). Then prove that $$2\sqrt{\left(\sum_{cyc}{x_{1}}\right)\left(\sum_{cyc}{\dfrac{1}{x_{1}}}\right)}-\sqrt{\left(\sum_{i=1}^{n-1}{x_{i}}\right)\left(\sum_{j=2}^{n}{\dfrac{1}{x_{j}}}\right)}\geq 2\sqrt{2}+1+4\left(k-1\right)$$