this inequality is homogeneous so we can put $a+b+c=1$ and apply Jensen's inequality for $f(x)=\frac{1}{\sqrt{x}}$ so we have : $\frac{a}{\sqrt{a^2+8bc}}+\frac{b}{\sqrt{b^2+8ac}}+\frac{b}{\sqrt{b^2+8ac}} \geq \frac{1}{\sqrt{a^3+b^3+c^3+24abc}}$ but : $1=(a+b+c)^3=a^3+b^3+c^3+6abc+3(a^2b+a^2c+b^2a+b^2c+c^2a+c^2b) \geq a^3+b^3+c^3+24abc$ , so indeed $\frac{1}{\sqrt{a^3+b^3+c^3+24abc}} \geq 1$ .

Sorry for reviving this post, firstly. I had a question about this inequality. Why does $a+b+c = 1$? Why can we do this?

Because the inequality is homogenous. Let's say $a+b+c=k$ Taking $a'=a/k, b'=b/k, c'=c/k$ would give the same result but $a'+b'+c'=1$

Rafal wrote: this inequality is homogeneous so we can put $a+b+c=1$ and apply Jensen's inequality for $f(x)=\frac{1}{\sqrt{x}}$ so we have : $\frac{a}{\sqrt{a^{2}+8bc}}+\frac{b}{\sqrt{b^{2}+8ac}}+\frac{b}{\sqrt{b^{2}+8ac}}\geq \frac{1}{\sqrt{a^{3}+b^{3}+c^{3}+24abc}}$ but : $1=(a+b+c)^{3}=a^{3}+b^{3}+c^{3}+6abc+3(a^{2}b+a^{2}c+b^{2}a+b^{2}c+c^{2}a+c^{2}b) \geq a^{3}+b^{3}+c^{3}+24abc$ , so indeed $\frac{1}{\sqrt{a^{3}+b^{3}+c^{3}+24abc}}\geq 1$ . I don't exactly understand how you are applying Jensen's.

i think the problem can be solved by using UCT by using UCT ,we need to prove that $\frac{a}{\sqrt{a^{2}+8bc}}\geq \frac{a^{\frac{4}{3}}}{a^{\frac{4}{3}}+b^{\frac{4}{3}}+c^{\frac{4}{3}}}$ it follows that $(2a^{\frac{4}{3}}+b^{\frac{4}{3}}+c^{\frac{4}{3}})(b^{\frac{4}{3}}+c^{\frac{4}{3}}) \geq 8a^{\frac{2}{3}bc}$ writing two similar inequalities of b and c and adding them up , we obtain the desire result

@Phelpedo: $\text{LHS}=af(a^{2}+8bc)+bf(b^{2}+8ca)+cf(c^{2}+8ab)$, that is all because $a+b+c=1$.

i have an other idea : we can put, $e^{x}=\frac{bc}{a^{2}}$, $e^{y}=\frac{ac}{b^{2}}$ and $e^{z}=\frac{ab}{c^{2}}$to get the inequality we have only to apply Jensen to the function $f(x)=\frac{1}{1+e^{x}}$ with $\lambda_{1}=\lambda_{2}=\lambda_{3}=\frac{1}{3}$ but i'm not sure of the convexity of $f$

HTA wrote: i think the problem can be solved by using UCT by using UCT ,we need to prove that $\frac{a}{\sqrt{a^{2}+8bc}}\geq \frac{a^{\frac{4}{3}}}{a^{\frac{4}{3}}+b^{\frac{4}{3}}+c^{\frac{4}{3}}}$ it follows that $(2a^{\frac{4}{3}}+b^{\frac{4}{3}}+c^{\frac{4}{3}})(b^{\frac{4}{3}}+c^{\frac{4}{3}}) \geq 8a^{\frac{2}{3}bc}$ writing two similar inequalities of b and c and adding them up , we obtain the desire result What does UCT stand for?

i'have made a mistake th efunction $f(x)=\frac{1}{1+8e^{x}}$

That function $\frac{1}{1+8e^{x}}$ is only convex when $e^{x}> 1/4$. However, things being less than 1/4 does not give the inequality much room to escape! EDIT: If we have e^x <= 1/4 , then we can apply Jensen to the remaining two variables to obtain a one-variable inequality, which then dies to taking a derivative. So, this method *does* work, however ugly it is.

Another solution: Let $ x = \sqrt \frac {a^2}{a^2 + 8bc}, y = \sqrt \frac {b^2}{b^2 + 8ac}, z = \sqrt \frac {c^2}{c^2 + 8ab}$. clearly, $ \frac {1}{x^2} = 1 + \frac {8bc}{a^2}$, so $ ( \frac {1}{x^2} - 1 )( \frac {1}{y^2} - 1) (\frac {1}{z^2} - 1) = 512$. We wish to prove that $ x + y + z \geq 1$. Let's prove that when $ x + y + z = 1$, then $ ( \frac {1}{x^2} - 1 )( \frac {1}{y^2} - 1) (\frac {1}{z^2} - 1) \geq 512$. This is easy: $ ( \frac {1}{x^2} - 1 )( \frac {1}{y^2} - 1) (\frac {1}{z^2} - 1) = \frac { (1 - x)(1 + x)(1 - y)(1 + y)(1 - z)(1 + z) }{x^2 y^2 z^2} = \frac { (x + y)(y + z)(x + z)(2x + y + z)(2y + x + z)(2z + x + y)}{x^2 y^2 z^2} \geq \frac { 512 \sqrt {xy} \sqrt {yz} \sqrt {xz} \sqrt [4]{x^2 yz} \sqrt [4]{y^2 xz} \sqrt [4]{z^2 xy} }{x^2 y^2 z^2} = 512$, as required. This is sufficient, since for some $ a,b,c$, if $ ( \frac {1}{a^2} - 1 )( \frac {1}{b^2} - 1) (\frac {1}{c^2} - 1) = 512$ ... (*), letting $ \lambda a = x, \lambda b = y, \lambda c = z$, where $ x + y + z = 1$, then we have, $ ( \frac {1}{x^2} - 1 )( \frac {1}{y^2} - 1) (\frac {1}{z^2} - 1) \geq 512$ - then comparing with (*), we have $ \lambda \leq 1$, so $ a + b + c \geq 1$.

Here's another solution, quite a simple one. by Cauchy-Schwarz: \[ (\frac{a}{\sqrt{a^2 + 8bc}} + \frac{b}{\sqrt{b^2 + 8ca}} + \frac{c}{\sqrt{c^2 + 8ab}})^2 (\sum_{cyc}{a(a^2+8bc)})\] \[ \geq(a+b+c)^3\] And: \[ (a+b+c)^3\geq\sum_{cyc}{a(a^2+8bc)}\] Thus, we get: \[ (\frac{a}{\sqrt{a^2 + 8bc}} + \frac{b}{\sqrt{b^2 + 8ca}} + \frac{c}{\sqrt{c^2 + 8ab}})^2\] \[ \geq\frac{(a+b+c)^3}{(\sum{a(a^2+8bc)})}\] \[ \geq1\] $ Q.E.D.$

Because of the homogeneity we can take $ a.b.c = 1$. Inequality equivalent to $ \sqrt {\frac {a^3}{a^3 + 8}} + \sqrt {\frac {b^3}{b^3 + 8}} + \sqrt {\frac {c^3}{c^3 + 8}}\geq1$. Let $ a = \frac {2}{x} , b = \frac {2}{y} , c = \frac {2}{z}$ . $ \frac {1}{\sqrt {1 + x^3}} + \frac {1}{\sqrt {1 + y^3}} + \frac {1}{\sqrt {1 + z^3}}\geq1$ By the Arithmetic Geometric Mean Inequality $ \sqrt {(x + 1)(x^2 - x + 1)}\leq\frac {x^2 + 2}{2}$ and we will prove that $ \frac {2}{x^2 + 2} + \frac {2}{y^2 + 2} + \frac {2}{z^2 + 2}\geq1$ $ \Longleftrightarrow$ $ 2(x^2y^2 + y^2z^2 + x^2z^2) + 8(x^2 + y^2 + z^2) + 24\geq2(x^2y^2 + y^2z^2 + x^2z^2) + 4(x^2 + y^2 + z^2) + x^2y^2z^2$ $ \Longleftrightarrow$ $ x^2 + y^2 + z^2\geq12$ which is obviously true. ($ x.y.z=8$)

What does UCT stand for? If it means showing that each of three fractions is greater the f(a)/(f(a)+f(b)+f(c)), hence their sum must be greater then 1, then what is the motivation for choosing f(x)=$ x^{4/3}$?

Here is my AM-GM's proof: Clearly,the problem is equivalent to prove that : $ \sum \frac{1}{ \sqrt{8x+1} } \geq 1$ With $ x,y,z \geq 0$ and $ xyz=1$ Let $ 8x+1=a^2,8y+1=b^2$ and $ 8z+1=c^2$; We have to prove that : $ ab+bc+ca \geq abc$ with $ a,b,c \geq 1$ and $ (a^2-1)(b^2-1)(c^2-1)=512 \Leftrightarrow a^2b^2c^2-(a^2b^2+b^2c^2+c^2a^2)=513-(a^2+b^2+c^2)$; Squaring both sides,the inequality becomes: $ \sum a^2b^2 +2abc(a+b+c) \geq a^2b^2c^2$ Wich can be reduced to: $ a^2+b^2+c^2+2abc(a+b+c) \geq 513$ But this is clearly true because : $ abc = \sqrt{(8x+1)(8y+1)(8z+1)} \geq 27$ $ \sum a \geq 3 \sqrt[3]{abc} \geq 9$ $ \sum a^2 \geq \frac{ (\sum a)^2}{3} \geq 27$ Done !

orl wrote: Prove that for all positive real numbers $ a,b,c$, $ \frac {a}{\sqrt {a^2 + 8bc}} + \frac {b}{\sqrt {b^2 + 8ca}} + \frac {c}{\sqrt {c^2 + 8ab}} \geq 1.$ This problem is introduced as Example 2.2.1 on pp. 38-39 in new book : Ji Chen, Chao-Cheng Ji, Algebraic Inequalities (Mathematical Olympiad Proposition people Lecture), Shanghai Scientific & Technological Educational Publishing House, 2009, 225p. (ISBN 978-7-5428-4848-2/O·613) epitomy01 wrote: Let $ x = \sqrt \frac {a^2}{a^2 + 8bc}, y = \sqrt \frac {b^2}{b^2 + 8ca}, z = \sqrt \frac {c^2}{c^2 + 8ab},$ clearly, $ \frac {1}{x^2} = 1 + \frac {8bc}{a^2},$ so $ \left(\frac {1}{x^2} - 1\right)\left(\frac {1}{y^2} - 1\right)\left(\frac {1}{z^2} - 1\right) = 512.$ We wish to prove that $ x + y + z \geq 1.$ $ (x + y + z - 1)\left[1 + x + y + z + 2(yz + zx + xy) + 18xyz + 19xyz(x + y + z) + 57xyz(yz + zx + xy)\right]$ $ = 512x^2y^2z^2 - \prod{\left(1 - x^2\right)} + \sum{\frac {x}{2}(y - z)^2(4 + x + 19yz + 114xyz)}\geq0.$ See also : http://www.mathlinks.ro/viewtopic.php?t=118059 crazyfehmy wrote: $ \frac {1}{\sqrt {1 + x^3}} + \frac {1}{\sqrt {1 + y^3}} + \frac {1}{\sqrt {1 + z^3}}\geq1.$ By the Arithmetic Geometric Mean Inequality $ \sqrt {(x + 1)(x^2 - x + 1)}\leq\frac {x^2 + 2}{2}$ and we will prove that $ \frac {2}{x^2 + 2} + \frac {2}{y^2 + 2} + \frac {2}{z^2 + 2}\geq1.$ ($ xyz = 8$) The last problem is introduced as Example 2.2.12 on page 47 in the above book. See also : http://www.mathlinks.ro/viewtopic.php?t=32332

Thank you very mych....

\[ \frac{a}{\sqrt{a^{2}+8bc}}+\frac{b}{\sqrt{b^{2}+8ca}}+\frac{c}{\sqrt{c^{2}+8ab}}\geq 1 \] \[\sqrt{a^{2}+(2b)^{2}+(2c)^{2}}\geq \sqrt{a^{2}+8bc} \] similarly for other \[\sum {\frac{a}{\sqrt{a^{2}+8bc}}}\geq \sum{\frac{a}{a^{2}+(2b)^{2}+(2c)^{2}}}\] $X=\frac{b^{2}+c^{2}}{a^{2}}$ similarly for other \[\sum {\frac{1}{\sqrt{1+4X}}}\geq 1\] \[\sum{X}=2\] you know by AM HM inequality \[{\sum {\sqrt{1+4X}}}\sum {\frac{1}{\sqrt{1+4X}}}\geq 9\] then \[\sum{\frac{1}{\sqrt{1+4X}}}\geq \frac{9}{\sum{\sqrt{1+4X}}}\] then by RMS $\geq$ AM inequality %\[\sqrt{\frac{\sum{(1+4X)}}{3}} \geq \frac{\sum{\sqrt{(1+4X)}}{3}\] \[\frac{3}{\sum{(1+4X)}} \geq \sqrt{\frac{3}{\sum{(1+4X)}}}\] above rhs of inequality becomes \[\geq 3\sqrt{\frac{3}{3+4\sum{X}}}\] you know that \[\sum{X}=2\] \[\geq \sqrt{\frac{27}{11}} \] \[\geq 1.\] finally proved

in previous message RMS \[\geq\] AM i have forget about raising square root to 1+4X so equation is like this \[\frac{3}{\sum{\sqrt{(1+4X)}}} \geq \sqrt{\frac{3}{\sum{(1+4X)}}}\] this is wrong \[ \frac{3}{\sum{(1+4X)}}\geq\sqrt{\frac{3}{\sum{(1+4X)}}} \]

Generalization 1 Let $a,b,c,p,\lambda$ be positive reals. Then prove that $$\dfrac{a^p}{\sqrt{a^{2p}+\lambda bc}}+\dfrac{b^p}{\sqrt{b^{2p}+\lambda ca}}+\dfrac{c^p}{\sqrt{c^{2p}+\lambda ab}}\geq \dfrac{\lambda ^2}{4\left(3\sqrt[3]{\lambda ^2}+4\right)}$$

We have the following by Holder's: \[\left(\sum_{cyc}\frac{a}{\sqrt{a^2+8bc}}\right)^2\left(\sum_{cyc}a(a^2+8bc)\right)\ge (a+b+c)^3\]so it suffices to prove \[(a+b+c)^3\ge \sum_{cyc}a(a^2+8bc)\]equivalently \[a^2b+ab^2+bc^2+b^2c+c^2a+ca^2\ge 6abc\]which is true by direct AM-GM. $\square$

ehuseyinyigit wrote: Generalization 1 Let $a,b,c,p,\lambda$ be positive reals. Then prove that $$\dfrac{a^p}{\sqrt{a^{2p}+\lambda bc}}+\dfrac{b^p}{\sqrt{b^{2p}+\lambda ca}}+\dfrac{c^p}{\sqrt{c^{2p}+\lambda ab}}\geq \dfrac{\lambda ^2}{4\left(3\sqrt[3]{\lambda ^2}+4\right)}$$ Tangent line or holder

Bryan0224 wrote: Tangent line or holder Yes, it has a proof by Holder's Inequality.

Generalization 2 For any $a,b,c,\lambda$ ($\lambda \geq 8$) positive reals, the inequality $$\dfrac{a}{\sqrt{a^{2}+\lambda bc}}+\dfrac{b}{\sqrt{b^{2}+\lambda ca}}+\dfrac{c}{\sqrt{c^{2}+\lambda ab}}\geq 1-\dfrac{\left(\sqrt[3]{\lambda ^2}-4\right)\left(\sqrt[3]{\lambda^2}+2\right)^2}{6\sqrt[3]{\lambda}\left(\lambda +2\sqrt[3]{\lambda}\right)+8}$$ holds.

ehuseyinyigit wrote: Generalization 2 For any $a,b,c,\lambda$ ($\lambda \geq 8$) positive reals, the inequality $$\dfrac{a}{\sqrt{a^{2}+\lambda bc}}+\dfrac{b}{\sqrt{b^{2}+\lambda ca}}+\dfrac{c}{\sqrt{c^{2}+\lambda ab}}\geq 1-\dfrac{\left(\sqrt[3]{\lambda ^2}-4\right)\left(\sqrt[3]{\lambda^2}+2\right)^2}{6\sqrt[3]{\lambda}\left(\lambda +2\sqrt[3]{\lambda}\right)+8}$$ holds. Because :$$\frac{3}{\sqrt{1+\lambda}}\ge1-\dfrac{\left(\sqrt[3]{\lambda ^2}-4\right)\left(\sqrt[3]{\lambda^2}+2\right)^2}{6\sqrt[3]{\lambda}\left(\lambda +2\sqrt[3]{\lambda}\right)+8},\left(\lambda \geq 8\right).$$ .

Nice! Here was the $n$-var extension (GENERALIZATION 3) :

By Holder, $$\left( \sum_{cyc} \frac{a}{\sqrt{a^2 + 8bc}} \right)^2 \left( \sum_{cyc} a(a^2+8bc) \right) \geq (a+b+c)^3$$so it suffices to show that $a^3+b^3+c^3+24abc \leq (a+b+c)^3$, or $\sum_{sym} a^2b \geq 6abc$, which is trivial by Muirhead.

huh.... $$\left( \sum_{cyc} \frac{a}{\sqrt{a^2 + 8bc}} \right)^2 \left( \sum_{cyc} a(a^2+8bc) \right) \geq (a+b+c)^3$$Lemma:$(a+b+c)^3\geq a^3+b^3+c^3+24abc$ Proof: $$(a+b+c)^3=a^3+b^3+c^3+3\left(\sum_{sym} a^2b\right) +6abc \ge24abc+a^3+b^3+c^3\ \iff \sum_{sym} a^2b\ge 6abc $$which is true by AM-GM and we are done $\quad\square$

My first ever problem to post in this platform! Okay so we have to show that \[ \sum_{\text{cyc}} \frac{a}{\sqrt{a^2 + 8bc}} \geq 1. \] We can easily prove this by applying Hölder's inequality carefully. Let us choose \( p = 2 \) and \( q =1 \). Applying Hölder's gives us: \[ \sum_{\text{cyc}} \left( \frac{a}{\sqrt{a^2 + 8bc}} \right)^2 \cdot (a(a^2 + 8bc) \geq (a+b+c)^3. \] If we expand this, we get: \[ \sum_{\text{cyc}} \frac{a^2}{a^2 + 8bc} \geq \frac{(a+b+c)^3}{\sum_{\text{cyc}} a(a^2 + 8bc)}. \] Thus, it suffices to show that \[ \frac{(a+b+c)^3}{\sum_{\text{cyc}} a(a^2 + 8bc)} \geq 1, \] which is equivalent to proving \[ (a+b+c)^3 \geq \sum_{\text{cyc}} a(a^2 + 8bc). \] Now, we compute \[ \sum_{\text{cyc}} a(a^2 + 8bc) = \sum_{\text{cyc}} (a^3 + 8abc) = a^3 + b^3 + c^3 + 24abc. \] So we need to show that \[ (a+b+c)^3 \geq a^3 + b^3 + c^3 + 24abc. \] Using the expansion, we have: \[ (a+b+c)^3 = a^3 + b^3 + c^3 + 3(a+b)(b+c)(c+a). \] It suffices to prove that \[ 3(a+b)(b+c)(c+a) \geq 24abc. \] By the AM-GM inequality, we can derive: \[ a^2b + a^2c + ab^2 + ac^2 + b^2c + bc^2 \geq 6abc. \] We can show this using AM-GM as follows: 1. For \( a^2b + b^2c + c^2a \): \[ a^2b + b^2c + c^2a \geq 3 \sqrt[3]{(a^2b)(b^2c)(c^2a)} = 3abc. \] 2. For \( ab^2 + bc^2 + ca^2 \): \[ ab^2 + bc^2 + ca^2 \geq 3 \sqrt[3]{(ab^2)(bc^2)(ca^2)} = 3abc. \] Adding these inequalities gives: \[ a^2b + a^2c + ab^2 + ac^2 + b^2c + bc^2 \geq 6abc. \] Thus, we conclude that \[ \sum_{\text{cyc}} \frac{a}{\sqrt{a^2 + 8bc}} \geq 1, \] and the proof is complete.

Claim: $\sum_{cyc} a(a^2 + 8bc) \le (a+b+c)^3$. Proof: Expand to get Muirhead's inequality. $\square$ Now, by Holder's inequality, $$\left(\frac{a}{\sqrt{a^2 + 8bc}} + \frac{b}{\sqrt{b^2 + 8ca}} + \frac{c}{\sqrt{c^2 + 8ab}}\right)^2 \left((a+b+c\right)^3)$$$$ \ge \left(\frac{a}{\sqrt{a^2 + 8bc}} + \frac{b}{\sqrt{b^2 + 8ca}} + \frac{c}{\sqrt{c^2 + 8ab}}\right)^2 \left(\sum_{cyc} a(a^2 + 8bc) \right)$$$$\ge \left(\sum_{cyc} {a}\right)^3,$$so we're done.

I have discussed this problem as an application of Holder in my inequality tutorial playlist. Here is the video

It’s pretty simple. Just apply Cauchy schwarz then am gm kills the problem

By Holder's Inequality, $$\text{LHS}^2 \cdot \left( \sum_{cyc} a(a^2+8bc) \right) \geq (a+b+c)^3,$$so it suffices to show that $$a^3+b^3+c^3+24abc \leq (a+b+c)^3$$which is well-known. (for a complete proof though, note we can fully expand everything and it will reduce to Muirhead.) QED