nice it is equivalent to: $ \sum_{cyc}\frac{a}{\sqrt{a(b+c)}}\geq 2\sqrt{\frac{(a+b+c)(ab+bc+ca)}{(a+b)(b+c)(c+a)}}$ now using Jensen on $ f(x)=\frac{1}{\sqrt{x}}$ we get $ \sum_{cyc}\frac{a}{\sqrt{a(b+c)}}\geq \frac{a+b+c}{\sqrt{\frac{\sum_{sym} a^2b}{a+b+c}}}$ so all we need to prove is $ (a+b+c)^2(\sum_{sym} a^2b+2abc)\geq 4(ab+bc+ca)(\sum_{sym} a^2b)$ now let $ c$ be the smallest number among $ a,b,c$ and we see we can rewrite the above as $ (a-b)^2(a^2b+b^2a+a^2c+b^2c-ac^2-bc^2)+c^2(a+b)(c-a)(c-b)\geq 0$

the inequality is equivalent to \[ \sum \sqrt{a(a+b)(a+c)}\geq 2\sqrt{(a+b+c)(ab+bc+ca)}\] after squaring both sides and cancelling some terms we have that it is equivalent to \[ \sum a^3+abc+2(b+c)\sqrt{bc(a+b)(a+c)}\geq \sum 3a^2b+3a^2c+4abc\] from schur's inequality we have that it is enough to prove that \[ \sum (b+c)\sqrt{(ab+b^2)(ac+c^2)}\geq \sum a^2b+a^2c+2abc\] from cauchy-schwarz we have that $ \sqrt{(ab+b^2)(ac+c^2)}\geq a\sqrt{bc}+bc$, so we have that \[ \sum (b+c)\sqrt{(ab+b^2)(ac+c^2)}\geq \sum a(b+c)\sqrt{bc}+bc(b+c)\geq \sum a^2b+a^2c+2abc\] as we wanted to prove.

the same idea as campos , but a little bit different squaring the both sides , our ine is equivalent to: $ \sum{a^3}-3\sum{ab(a+b)}-9abc+2\sum{\sqrt{a(a+b)(a+c)}\sqrt{b(b+c)(b+a)} \geq 0}$ but : $ a(a+b)(a+c)\b(b+c)(b+a) = (a^3+a^2c+a^2b+abc)(ab^2+b^2c+b^3+abc) \geq (a^2b+abc+ab^2+abc)^2$ (cauchy) so we want to prove that: $ a^3+b^3+c^3-ab(a+b)-ac(a+c)-bc(b+c)+3abc\geq 0$ , which is schur

can we let tan x=1/a tan y=1/b tan z=1/c (cosx/sinx^3)^1/2+(cosy/siny^3)^1/2+(cosz/sinz^3)>2(cotx+coty+cotz)^1/2? i think this idea and continue this is too complex but i wnt to know this idea will work or not?

campos wrote: the inequality is equivalent to \[ \sum \sqrt {a(a + b)(a + c)}\geq 2\sqrt {(a + b + c)(ab + bc + ca)} \] I was wondering... is the function $ f(x)=\sqrt{(x+a)(x+b)(x+c)}$ convex for $ a,b,c>0$? Then, here we could use Jensen after multiplying both sides by $ \sqrt{2}$. (This was on a MOP team contest this year, and another student proved this inequality assuming that function is convex...)

campos wrote: Quote: the inequality is equivalent to \sum \sqrt {a(a + b)(a + c)}\geq 2\sqrt {(a + b + c)(ab + bc + ca)} Where from you derived this new inequality???

note that $ a^3+a=a(a^2+(ab+bc+ca))=a(a+b)(a+c)$, and the rhs is obtained by multiplying the original by $ ab+bc+ca=1$...

Why don't we use Holder's Inequality?

Quote: Why don't we use Holder's Inequality? How? Anyway i almost finished this problem,just to proof that: sqrt{a/b+c} +sqrt { b/a+c} +sqrt { c/a+b} >= 3/sqrt(2) I think i saw this inequality on mathlinks but i can't remember where... If somebody knows proof or where i can find it please write it....

this inequality isn't true, take for example $ a=b$ and $ c\rightarrow 0$... the one that is a well-known inequality is $ \sum \sqrt{\dfrac{a}{b+c}}\geq 2$...

Omid Hatami wrote: Let $ a,b,c > 0$ and $ ab + ac + bc = 1$. Prove that: \[ \sqrt {a^3 + a} + \sqrt {b^3 + b} + \sqrt {c^3 + c}\geq2\sqrt {a + b + c} \] See my solution here: http://mathvn.org/forum/viewthread.php?thread_id=125

@ can_hang2007 i understand your solution accept the last inequality (how to solve it ), and i find its equivalent to T[4,1,0]+T[2,2,1] >= T[3,2,0]+T[3,1,1] but i can't prove it @campos at yours "well-known inequality" have you tried putting a=b=c ? Cause then you get 3 /sqrt(2)>=2 thus it's not true... So i think inequality i posted Quote: sqrt{a/b+c} +sqrt { b/a+c} +sqrt { c/a+b} >= 3/sqrt(2) is true...

Smilley wrote: @ can_hang2007 i understand your solution accept the last inequality (how to solve it ), and i find its equivalent to T[4,1,0]+T[2,2,1] >= T[3,2,0]+T[3,1,1] but i can't prove it You can prove it easily by SOS or pqr method. Besides the stronger is more interesting: $ \frac {a^2 + b^2 + c^2}{ab + bc + ca} + 2\sum\ \frac {ab}{(a + b)^2} \geq\ \frac {5}{2}$ I have had 2 proofs for this stronger by SOS

$ \dfrac{3}{\sqrt 2}>2$

nguoivn wrote: Smilley wrote: @ can_hang2007 i understand your solution accept the last inequality (how to solve it ), and i find its equivalent to T[4,1,0]+T[2,2,1] >= T[3,2,0]+T[3,1,1] but i can't prove it You can prove it easily by SOS or pqr method. Besides the stronger is more interesting: $ \frac {a^2 + b^2 + c^2}{ab + bc + ca} + 2\sum\ \frac {ab}{(a + b)^2} \geq\ \frac {5}{2}$ I have had 2 proofs for this stronger by SOS How is that inequality connected to the one posted originally? Why is it stronger?

campos wrote: $ \dfrac{3}{\sqrt 2} > 2$ I don't understand your idea, my friend Mathias_DK wrote: How is that inequality connected to the one posted originally? Why is it stronger? Because of an old result of can_hang and me (also the problem which somebody proposed) $ \frac{8abc}{(a+b)(b+c)(c+a)} \geq\ 2 \sum\ \frac{ab}{(a+b)^2} - \frac{1}{2}$ (This ineq is trues with all the reals $ a, b, c$ and we can prove it easily by middle school knowledge)

@campos yes of course you're right i made a mistake @nguoivn I've heard for SOS method, but what is pqr method ? Do you have some script about this method,it will be helpful to me...

Here is

Smilley wrote: @nguoivn I've heard for SOS method, but what is pqr method ? Do you have some script about this method,it will be helpful to me... You can see here: http://www.maths.vn/forums/showthread.php?t=11365

prove(吴国盛) $a^3+a=a(a^2+1)=a(a^2+ab+bc+ca)$ $b^3+b=b(b^2+1)=b(b^2+ab+bc+ca)$ $c^3+c=c(c^2+1)=a(a^2+ab+bc+ca)$ By C-S,have $ (\sqrt {a^3 + a} + \sqrt {b^3 + b} + \sqrt {c^3 + c})^2\le (a+b+c)(a^2+b^2+c^2+3(ab+bc+ca))$ $\le (a+b+c)((a+b+c)^2+\frac{1}{3}(a+b+c)^2)$ $=\frac{2\sqrt{3}}{3}\sqrt {(a + b + c)^3}$

In triangle,have this: $\frac{1}{6}\,{\frac {\sqrt {3} \left( a-b+c \right) \left( a+b-c \right) \left( a+b \right) \left( c+a \right) bc}{{r}^{2} \left( {s}^{4}+ \left( 32\,{R}^{2}+4\,Rr+2\,{r}^{2} \right) {s}^{2}+r \left( 4\,R+r \right) ^{3} \right) }}\geq \sqrt{\frac{\tan{\frac{A}{2}}(\tan^2{\frac{A}{2}}+1)}{(\sum{\tan{\frac{A}{2}}}^3}}$

xzlbq wrote: Omid Hatami wrote: Let $a,b,c > 0$ and $ab+bc+ca = 1$. Prove that: \[ \sqrt {a^3 + a} + \sqrt {b^3 + b} + \sqrt {c^3 + c}\geq2\sqrt {a + b + c}. \] Let $a,b,c > 0$ and $ab+bc+ca = 1$. Prove that: \[ \sqrt {a^3 + a} + \sqrt {b^3 + b} + \sqrt {c^3 + c}\leq\frac{2\sqrt{3}}{3}\sqrt {(a + b + c)^3}. \] stronger Let $a,b,c > 0$ and $ab+bc+ca = 1$. Prove that: \[\sqrt { \left( a+b+c \right) \left( \left( a+b+c \right) ^{2}+1 \right) }\geq \sqrt {{a}^{3}+a}+\sqrt {{b}^{3}+b}+\sqrt {{c}^{3}+c}\]

Let $a,b,c,d > 0$ and $ab+bc+cd+ad=1$. Prove that: \[\sqrt {{a}^{3}+a}+\sqrt {{b}^{3}+b}+\sqrt {{c}^{3}+c}+\sqrt {{d}^{3}+d }\geq 2\,\sqrt {1+4\,abcd}\sqrt {a+b+c+d} \]

xzlbq wrote: xzlbq wrote: Omid Hatami wrote: Let $a,b,c > 0$ and $ab+bc+ca = 1$. Prove that: \[ \sqrt {a^3 + a} + \sqrt {b^3 + b} + \sqrt {c^3 + c}\geq2\sqrt {a + b + c}. \] Let $a,b,c > 0$ and $ab+bc+ca = 1$. Prove that: \[ \sqrt {a^3 + a} + \sqrt {b^3 + b} + \sqrt {c^3 + c}\leq\frac{2\sqrt{3}}{3}\sqrt {(a + b + c)^3}. \] stronger Let $a,b,c > 0$ and $ab+bc+ca = 1$. Prove that: \[\sqrt { \left( a+b+c \right) \left( \left( a+b+c \right) ^{2}+1 \right) }\geq \sqrt {{a}^{3}+a}+\sqrt {{b}^{3}+b}+\sqrt {{c}^{3}+c}\] From C-S we have: $$(\sum a)(\sum a^2+3\sum ab)\ge(\sum\sqrt{a(a^2+\sum ab)})^2$$Then from problem we know: $$\sum\sqrt{a^3+a}=\sum\sqrt{a(a^2+\sum ab)}$$Hence we need to prove: $$(\sqrt{(\sum a)((\sum a)^2+1))})^2\ge(\sum a)(\sum a^2+3\sum ab)$$Or $$(\sum a)((\sum a)^2+1)\ge(\sum a)(\sum a^2+3ab)$$Or $$\sum a^2+2ab+1\ge\sum a^2+3\sum ab$$Or $$1=\sum ab\ge\sum ab$$So we're done...!

Omid Hatami wrote: Let $a,b,c > 0$ and $ab+bc+ca = 1$. Prove that: \[ \sqrt {a^3 + a} + \sqrt {b^3 + b} + \sqrt {c^3 + c}\geq2\sqrt {a + b + c}. \] Let $a,b,c > 0$ and $ab+bc+ca = 1$. Prove that: \[\sum{\frac{1}{\sqrt{a^3+a}}}\geq \frac{3\sqrt{3}}{2}\sqrt{a+b+c}\]

xzlbq wrote: Omid Hatami wrote: Let $a,b,c > 0$ and $ab+bc+ca = 1$. Prove that: \[ \sqrt {a^3 + a} + \sqrt {b^3 + b} + \sqrt {c^3 + c}\geq2\sqrt {a + b + c}. \] Let $a,b,c > 0$ and $ab+bc+ca = 1$. Prove that: \[\sum{\frac{1}{\sqrt{a^3+a}}}\geq \frac{3\sqrt{3}}{2}\sqrt{a+b+c}\] Holder of course!

In triangle, \[{\frac {1}{\sqrt {\tan \left( 1/2\,A \right) \left( \left( \tan \left( 1/2\,A \right) \right) ^{2}+1 \right) }\sqrt {\tan \left( 1/2 \,A \right) +\tan \left( 1/2\,B \right) +\tan \left( 1/2\,C \right) }} }\geq \frac{3}{4}\,{\frac {\sqrt {3} \left( b+c-a \right) \left( a+b \right) \left( c+a \right) }{s \left( {s}^{2}+5\,{r}^{2}+8\,Rr \right) }}\]

see https://artofproblemsolving.com/community/c6t243f6h1653344_hard_inequality

Omid Hatami wrote: Let $a,b,c > 0$ and $ab+bc+ca = 1$. Prove that: \[ \sqrt {a^3 + a} + \sqrt {b^3 + b} + \sqrt {c^3 + c}\geq2\sqrt {a + b + c}. \] See also here: https://artofproblemsolving.com/community/c6h145158p823429

luofangxiang wrote: Can someone please tell why the inequality $$\sum_{\text{cyc}} \sqrt{ a^2 \sum_{\text{cyc}} a + abc } \ge \sqrt{ \left( \sum_{ \text{cyc}} a \right)^3 + 9abc}$$holds true ?

Let $a,b,c > 0$ and $a + b + c= 1$. Prove that $$ \sqrt {a^3 + a} + \sqrt {b^3 + b} + \sqrt {c^3 + c}\geq\sqrt {10(ab+bc+ca) }$$Let $a,b,c > 0$ and $\sqrt{a} + \sqrt{b} + \sqrt{c} = 3.$ Prove that $$ \sqrt{a^3+a} + \sqrt{b^3+b} + \sqrt{c^3+c}\geq\sqrt {6(ab+bc+ca) }$$

Notice that our inequality can be rewritten as \[\sum \sqrt{a^3+a(ab+bc+ca)} = \sum \sqrt{a(a+b)(a+c)} \ge 2 \sqrt{(a+b+c)(ab+bc+ca)}\]\[\iff 2 \sum (a+b)\sqrt{(a^2+ac)(b^2+bc)} + (a^3+b^3+c^3+3abc) + \sum a^2b \ge 4 \left(3abc+\sum a^2b\right).\] This inequality holds true, as \[a^3+b^3+c^3+3abc \ge \sum a^2b, \quad \text{and}\]\[\sum (a+b)\sqrt{(a^2+ac)(b^2+bc)} \ge \sum (a+b)(ab+c\sqrt{ab}) \ge \sum (a^2b+ab^2+2\sqrt{ab} \cdot c\sqrt{ab}) = 6abc + 2\sum a^2b. \quad \blacksquare\]

All nice solve

Solved with MathLuis. This was a pretty exciting inequality, but still really standard. Typical homogenize and done. We start off by homogenizing the given inequality as follows to get rid of the weird $ab+bc+ca=1$ condition. \[ \sqrt {a^3 + a(ab+bc+ca)} + \sqrt {b^3 + b(ab+bc+ca)} + \sqrt {c^3 + c(ab+bc+ca)}\geq 2\sqrt {(a + b + c)(ab+bc+ca)}. \]Note that by Holder's Inequality we have, \[\sqrt{a^3+a^2b+a^2c+abc} \cdot \sqrt{ab^2+b^3 + b^2c + abc} \ge a^2b + ab^2 + 2abc\]and also by Schur's Inequality \[a^3+b^3+c^3 +3abc \ge \sum_{cyc}a^2(b +c)\]Thus, \[\sum_{cyc}a^3 + 3abc + \sum_{cyc}a^2(b +c) + 2\sum_{cyc}\sqrt{a^3+a^2b+a^2c+abc} \cdot \sqrt{ab^2+b^3 + b^2c + abc} \ge 4 \sum_{cyc}a^2(b +c) + 12abc \]which implies that, \begin{align*} \left(\sqrt {a^3 + a(ab+bc+ca)} + \sqrt {b^3 + b(ab+bc+ca)} + \sqrt {c^3 + c(ab+bc+ca)}\right)^2 &\ge 4(a+b+c)(ab+bc+ca)\\ \sqrt {a^3 + a(ab+bc+ca)} + \sqrt {b^3 + b(ab+bc+ca)} + \sqrt {c^3 + c(ab+bc+ca)} &\ge 2\sqrt{(a+b+c)(ab+bc+ca)} \end{align*}which is precisely what we set out to show.