Note

Ok . I have a very nice proof for it . An excellent problem .

Maximal value is quite easy to find... Use Cauchy Schwarz on the sequences: $(a+b+c)((b+1)+(c+1)+(a+1)) \ge (\sqrt{a(b+1)}+ \sqrt{b(c+1)} + \sqrt{c(a+1)})^2$ There the above exp is less than equal to $3\sqrt{2}$

$\sum_{cyc}\sqrt{a(b+1)}=\frac{1}{\sqrt{2}}\sum_{cyc}\sqrt{2a(b+1)}\leq\frac{1}{\sqrt{2}}\sum_{cyc}\frac{2a+(b+1)}{2}=3\sqrt{2}.$ Equality holds when $a=b=c=1.$ KöMaL

Yes, but how do you find the minimum? That is the hard part.

If $a,b,c\in[0,2]$ and $a+b+c=3$. Then$$\sqrt{a(b+1)}+\sqrt{b(c+1)}+\sqrt{c(a+1)}\geq3.$$Equality holds when $a=2,b=1,c=0$ or $a=0,b=2,c=1$ or $a=1,b=0,c=2.$

From $a+b+c=3$ it follows that \begin{align*} \sum_{\text{cyc}}\sqrt{a(b+1)} &\ge{}\sum_{\text{cyc}}ab(b+1)(c+1) \\ &\ge{}3. \end{align*}

It follows . But wewe are your proofs ?

randomusername wrote: Suppose $a,b,c\in[0,2]$ and $a+b+c=3$. Find the maximal and minimal value of the expression $$\sqrt{a(b+1)}+\sqrt{b(c+1)}+\sqrt{c(a+1)}.$$ By rearrangmenent and Jensen we have: $\sqrt{a(b+1)}+\sqrt{b(c+1)}+\sqrt{c(a+1)} \leq \sqrt{a(a+1)}+\sqrt{b(b+1)}+\sqrt{c(c+1)} \leq 3\sqrt{\frac{a+b+c}{3}(\frac{a+b+c}{3}+1)}=3\sqrt{2}$ Equality holds when $a=b=c=1$

randomusername wrote: Suppose $a,b,c\in[0,2]$ and $a+b+c=3$. Find the maximal and minimal value of the expression $$\sqrt{a(b+1)}+\sqrt{b(c+1)}+\sqrt{c(a+1)}.$$ Here is my nice solution: $\bigstar$ By AM-GM we have $\sqrt{2a(b+1)}+\sqrt{2b(c+1)}+\sqrt{2c(a+1)}\le\frac{2a+b+1}{2}+\frac{2c+a+1}{2}+\frac{2c+a+1}{2}=6$,so $$\sqrt{a(b+1)}+\sqrt{b(c+1)}+\sqrt{c(a+1)}\le 3\sqrt{2}.$$Equality holds for $(a,b,c)=(1,1,1)$. $\bigstar$ We will prove that $\sqrt{a(b+1)}+\sqrt{b(c+1)}+\sqrt{c(a+1)}\ge 3$.By squaring that we obtain $$ab+bc+ca+2\sqrt{ab(b+1)(c+1)}+2\sqrt{ac(a+1)(b+1)}+2\sqrt{bc(c+1)(a+1)}\ge 6.$$But \begin{align*} 2\sqrt{ab(b+1)(c+1)}+2\sqrt{ac(a+1)(b+1)}+2\sqrt{bc(c+1)(a+1)} & \ge 2\sqrt{ab(b+1)(c+1)+ac(a+1)(b+1)+bc(c+1)(a+1)} \\ & =2\sqrt{6abc+ab^2+bc^2+ca^2+ab+bc+ca} \\ & \ge 2\sqrt{ab^2+bc^2+ca^2+ab+bc+ca} \end{align*} so we suffice to prove that $$ab+bc+ca+2\sqrt{ab^2+bc^2+ca^2+ab+bc+ca}\ge 6.$$Now,since $a,b,c\in [0,2]$ we deduce that $$a(2-a)(2-c)\ge 0,$$$$b(2-b)(2-a)\ge 0,$$$$c(2-c)(2-b)\ge 0.$$By adding those inequalities we deduce that $a(2-a)(2-c)+b(2-b)(2-a)+c(2-c)(2-b)\ge 0$ i.e. $$12+ab^2+bc^2+ca^2\ge 2(a^2+b^2+c^2)+2(ab+bc+ca)=a^2+b^2+c^2+9 \implies$$$$\implies ab^2+bc^2+ca^2\ge a^2+b^2+c^2-3.$$Thus $$ab+bc+ca+2\sqrt{ab^2+bc^2+ca^2+ab+bc+ca}\ge ab+bc+ca+2\sqrt{a^2+b^2+c^2-3+ab+bc+ca}=ab+bc+ca+2\sqrt{6-(ab+bc+ca)}.$$Let $ab+bc+ca=x$.It's enough to prove that $x+2\sqrt{6-x}\ge 6$,which is equivalent to $4(6-x)\ge (6-x)^2$ or $8x\ge 12+x^2$. Since $a,b,c\in [0,2]$ we deduce that $(2-a)(2-b)(2-c)\ge 0$,which is equivalent to $2(ab+bc+ca)\ge 4+abc$.Thus $2x\ge 4+abc\ge 4$,implying that $x\ge 2$.Let $x=2+y,y\ge 0$.We have $8x\ge 12+x^2$ $\iff$ $4y\ge y^2$.But $$2+y=ab+bc+ca\le\frac{(a+b+c)^2}{3}=3$$,so $y\le 1$. Therefore $y\in [0,1]$,from where it follows that $4y\ge y^2$. We conclude that $\sqrt{a(b+1)}+\sqrt{b(c+1)}+\sqrt{c(a+1)}\ge 3$.Equality holds for $(a,b,c)=(0,2,1),(1,0,2),(2,1,0)$.

ok huricane how did you assume or guess at first that the given exp. will be >=3?

Because the minimum should be taken when two variables are at opposite borders of the domain, when this domain is finite, and when the equality of all variables gives maximum.

(http://www.artofproblemsolving.com/community/c6h1154302p5471428)

MathKnight16 wrote: From $a+b+c=3$ it follows that $\sum_{\text{cyc}}\sqrt{a(b+1)} \ge\sum_{\text{cyc}}ab(b+1)(c+1)$ It's wrong! Try $a=2$, $b=1$ and $c=0$.

Upper bound is easy,i will prove it for lower bound which is $3$. Squaring,$$2\sum\sqrt{ab(b+1)(c+1)}+\sum ab\ge 6$$We have that $\sum ab\ge 2$ indeed suppose $a$ is greatest.Then by php,$a\ge 1$.And $\sum ab\ge a(b+c)=3a-a^2\ge 2$ as it is eq to $(a-1)(a-2)\le 0$ which is true. so we are to prove $$\sum \sqrt{ab(b+1)(c+1)}\ge 2$$. squaring again are neglecting the $2xy$ type of terms in expansion of form $(x+y+z)^2$, $$\sum ab(b+1)(c+1)\ge 4$$Now WLOG assume $a$ is greatest so $(a+1)(b+1)\ge2$ and $(c+1)(a+1)\ge 2$ as $a\ge 1$. Also,$(b+1)(c+1)\ge b+c+1=4-a\ge2$ substituting these above we reduce the ineq to $$\sum ab\ge 2$$which is proved already.So done

randomusername wrote: Suppose $a,b,c\in[0,2]$ and $a+b+c=3$. Find the maximal and minimal value of the expression $$\sqrt{a(b+1)}+\sqrt{b(c+1)}+\sqrt{c(a+1)}.$$

luofangxiang wrote: randomusername wrote: Suppose $a,b,c\in[0,2]$ and $a+b+c=3$. Find the maximal and minimal value of the expression $$\sqrt{a(b+1)}+\sqrt{b(c+1)}+\sqrt{c(a+1)}.$$ http://www.artofproblemsolving.com/community/c6h1244259p6372987