Stronger inequality at http://www.mathlinks.ro/Forum/viewtopic.php?t=25778 . Darij

From Cauchy we have \[ \left( \frac{1}{a^{3}(b+c)}+\frac{1}{b^{3}(c+a)}+\frac{1}{c^{3}(a+b)}\right) \Big( a(b+c)+b(c+a)+c(a+b) \Big) \geq \left( \frac 1a +\frac 1b+ \frac 1c \right)^2 \] and as $\frac 1a + \frac 1b + \frac 1c = ab+bc+ca$ we obtain that \[ \left( \frac{1}{a^{3}(b+c)}+\frac{1}{b^{3}(c+a)}+\frac{1}{c^{3}(a+b)}\right) \geq \frac{ ab+bc+ca} 2 \geq \frac{\sqrt [3]{a^2b^2c^2}}{2} = \frac 32 \] which is what we wanted to prove.

We have abc=1 Therefore 1/(a^3)(b+c) = bc/(a^2)(b+c) From Cauchy we have : bc/(a^2)(b+c) + 1/4b + 1/4c = bc/(a^2)(b+c)+(b+c)/4bc >= 1/a Similarly we have P>= 1/2a + 1/2b +1/2c>=3/2 which is what we wanted to prove

hope this has not been forgotten! i came to a weird solution, but cool and nice! we want to prove there exists and $r$ such that $\frac{1}{a^3(b+c)}\geq \frac{3}{2}\frac{a^r}{a^r+b^r+c^r}$. take $b=c$ and use that $a=b^{-2}$, then we get to $\frac{b^6}{2b}\geq \frac{3}{2}\frac{b^{-2r}}{b^{-2r}+2b^r}$ or $b^5\geq \frac{3}{1+2b^{3r}}$ or $f(b)=2b^{3r+5}+b^5-3\geq 0$. Obviously, $f(1)=0$, so we want $f'(1)=0$, this is $f'(b)=2(3r+5)b^{3r+2}+5b^4$, so we want $2(3r+5)+5=0$, or $r=\frac{-5}{2}$. we want to prove that $\frac{1}{a^3(b+c)}\geq \frac{3}{2}\frac{a^\frac{-5}{2}}{a^\frac{-5}{2}+b^\frac{-5}{2}+c^\frac{-5}{2}}$. use that $a=\frac{1}{bc}$ and let $x=\sqrt{b}$ and $y=\sqrt{c}$. we want to prove that $\frac{x^6y^6}{x^2+y^2}\geq \frac{3}{2}\frac{x^5y^5}{x^5y^5+\frac{1}{x^5}+\frac{1}{y^5}}$ or (after a bit of simplifications) $2(x^{10}y^{10}+x^5+y^5)\geq 3x^4y^4(x^2+y^2)$ from am-gm and rearrangement we have that $2(x^{10}y^{10}+x^5+y^5)\geq (x^{10}y^{10}+2x^4y)+(x^{10}y^{10}+2xy^4)\geq 3x^4y^4(x^2+y^2)$ and we're done!

the proof is long and i'm too tired to rewrite it so i'll just note: use well known substition $a = \frac{x}{y}$ and analogically for $b$ and $c$, so the assumption that $abc = 1$ is satisfied, ok now we multiply everything, we have something like this $\sum \frac{(y^{3}x)(y^{3}z)}{(x^{3}z)(y^{3}z)+(x^{3}y)(y^{3}x)}$ then we prove it pretty same as Nesbitt soz lads... i thought this way worths a mention...

orl wrote: Let $ a,b,c$ be positive real numbers such that $ abc=1$. Valentin Vornicu wrote: \[ \frac{\sqrt [3]{a^{2}b^{2}c^{2}}}{2}= \frac{3}{2}\] ?

x^n+y^n=z^n wrote: orl wrote: Let $ a,b,c$ be positive real numbers such that $ abc=1$. Valentin Vornicu wrote: \[ \frac{\sqrt [3]{a^{2}b^{2}c^{2}}}{2}= \frac{3}{2}\] ? Valentin seemed to have made a typo, it should be ${ \frac{3}{2}\sqrt [3]{a^{2}b^{2}c^{2}}}= \frac{3}{2}$.

Hello. I am really sorry for bringing up such an old topic. I am really new to olympiad inequalities and was just curious as to how you guys would have come up with that solution with Cauchy. Did you instinctively know that you were looking for $ ab + bc + ac$ in some form and know that the expression $ a(b+c) + b(a+c) + c(a+b)$ would similarly give you this expression? Or did you narrow down the possible approaches (from C-S, AM-GM, finding some expression always $ \ge 3/2$ but $ \le$ the left hand side, etc.) and try applying them until something worked?

http://www.math.ust.hk/excalibur/v11_n1.pdf

I've noticed that the most common solution to this problem has not been posted so I will post it for completeness: Replace $a,b,c$ with $\frac{1}{x},\frac{1}{y},\frac{1}{z}$. We still have $xyz=1$. We aim to prove $\sum\frac{x^3yz}{y+z}\ge \frac{3}{2}$ or equivalently $\sum\frac{x^2}{y+z}\ge\frac{3}{2}$. From here just about any method will work. First by Cauchy-Schwarz: $\left(\sum y+z\right) \left(\sum\frac{x^2}{y+z}\right)\ge (x+y+z)^2$. This rearranges to $\sum\frac{x^2}{y+z}\ge \frac{x+y+z}{2}\ge\frac{3\sqrt[3]{xyz}}{2}=\frac{3}{2}$. Another way is using Chebyshev and Nesbitt: $3\left(\sum\frac{x^2}{y+z}\right)\ge\left(\sum\frac{x}{y+z}\right)(x+y+z)\ge\frac{9}{2}$.

WakeUp wrote: I've noticed that the most common solution to this problem has not been posted so I will post it for completeness: Replace $a,b,c$ with $\frac{1}{x},\frac{1}{y},\frac{1}{z}$. We still have $xyz=1$. We aim to prove $\sum\frac{x^3yz}{y+z}\ge \frac{3}{2}$ or equivalently $\sum\frac{x^2}{y+z}\ge\frac{3}{2}$. . From Cauchy: $\frac{x^2}{y+z}+\frac{y+z}{4}\ge x$

orl wrote: Let $ a$, $ b$, $ c$ be positive real numbers such that $ abc = 1$. Prove that \[ \frac {1}{a^{3}\left(b + c\right)} + \frac {1}{b^{3}\left(c + a\right)} + \frac {1}{c^{3}\left(a + b\right)}\geq \frac {3}{2}. \] let $a=\frac{x}{y}$, $b=\frac{y}{z}$ and $c=\frac{z}{x}$. Inequality becomes: $\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}$$\geq \frac{3}{2}$. Now, by $AM-GM$ inequality we have: $\frac{x^2}{y+z}+\frac{y+z}{4}\geq 2\sqrt{\frac{x^2}{y+z}.\frac{y+z}{4}}=x$

$ \sum{\frac{1}{a^{3}(b+c)}}=\sum{\frac{b^{2}c^{2}}{a(b+c)}\geq \frac{(\sum{bc})^{2}}{2\sum{bc}} }$ $ =\frac{\sum{bc}}{2}\geq \frac{3}{2} $

orl wrote: Let $ a$, $ b$, $ c$ be positive real numbers such that $ abc = 1$. Prove that \[ \frac {1}{a^{3}\left(b + c\right)} + \frac {1}{b^{3}\left(c + a\right)} + \frac {1}{c^{3}\left(a + b\right)}\geq \frac {3}{2}. \] \[ \frac {1}{a^{3}\left(b + c\right)} + \frac {1}{b^{3}\left(c + a\right)} + \frac {1}{c^{3}\left(a + b\right)}\] \[=\sum_{cyc} \frac{(\frac{1}{a^2})}{a(b+c)}\] \[\ge \frac{(\sum \frac{1}{a})^2}{2(ab+bc+ca)}\] (by Titu's or Cauchy-Schwarz inequality) \[ =\frac{ab+bc+ca}{2}\ge \frac{3}{2}\] by the AM-GM inequality. Edit: I believe these types of inequalities should not be given anymore.They all fall prey to Titu/Cauchy-Schwarz. This inequality here is better: I am posting this here too.please type new solutions there: nttu wrote: Let $ a, b, c >0 , abc =1 $ . Prove that : $ \frac{1}{a^3(b+c)} + \frac{1}{b^3(a+c)} + \frac{1}{c^3(b+a)} \geq \frac{3}{2} + \frac{1}{4}[\frac{a(c-b)^2}{b+c} + \frac{b(c-a)^2}{c+a} + \frac{c(b-a)^2}{a+b}] $

orl wrote: Let $ a$, $ b$, $ c$ be positive real numbers such that $ abc = 1$. Prove that \[ \frac {1}{a^{3}\left(b + c\right)} + \frac {1}{b^{3}\left(c + a\right)} + \frac {1}{c^{3}\left(a + b\right)}\geq \frac {3}{2}. \] $x=\frac{1}{a}$, $y=\frac{1}{b}$ dhe $z=\frac{1}{c}$ $a=\frac{1}{x}$, $b=\frac{1}{y}$, $c=\frac{1}{z}$ dhe $xyz=1.$ $LHS=\frac{1}{a^3(b+c)}+\frac{1}{b^3(c+a)}+\frac{1}{c^3(a+b)}=$ $=\frac{1}{\frac{1}{x^3}\left (\frac{1}{y}+\frac{1}{z} \right)}+\frac{1}{\frac{1}{y^3}\left (\frac{1}{z}+\frac{1}{x} \right)}+\frac{1}{\frac{1}{z^3}\left (\frac{1}{x}+\frac{1}{y} \right)}=$ $=\frac{x^3}{\frac{y+z}{yz}}+\frac{y^3}{\frac{z+x}{zx}}+\frac{z^3}{\frac{x+y}{xy}}=\frac{x^3yz}{y+z}+\frac{y^3zx}{z+x}+\frac{z^3xy}{x+y}=$ $=\frac{x^2xyz}{y+z}+\frac{y^2xyz}{z+x}+\frac{z^2xyz}{x+y}=\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}.$ We will prove that $S\geq \frac{3}{2}$, where $S=\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}$. This inequality we multiply with $[(b+c)+(c+a)+(a+b)]$ and we obtain: $[(y+z)+(z+x)+(x+y)]\cdot S=\left( \frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y} \right) [(y+z)+(z+x)+(x+y)]=$ $=x^2+y^2+z^2+ \left(x^2\frac{z+x}{y+z}+y^2\frac{y+z}{z+x} \right)+ \left(y^2\frac{x+y}{z+x}+z^2\frac{z+x}{x+y} \right)+ \left(z^2\frac{y+z}{x+y}+x^2\frac{x+y}{y+z} \right)\geq$ $\geq x^2+y^2+z^2+2\sqrt{x^2y^2}+2\sqrt{y^2z^2}+2\sqrt{z^2x^2}=x^2+y^2+z^2+2xy+2yz+2zx=(x+y+z)^2.$ From here $2(x+y+z)\cdot S\geq (x+y+z)^2$, we have $S\geq \frac{x+y+z}{2}.$ Finally, $S\geq \frac{x+y+z}{2}=\frac{3}{2}\cdot \frac{x+y+z}{3}\geq \frac{3}{2}\cdot \sqrt[3]{xyz}=\frac{3}{2}\cdot \sqrt[3]{1}=\frac{3}{2}.$ Equality holds for $x=y=z=1$, i.e. $a=b=c=1.$

My proof(1996):\[\sum ( \frac{1}{a^3(b+c)}+\frac{a(b+c)}{4}) \ge\sum bc,\] \[\sum \frac{1}{a^3(b+c)} \ge \frac{1}{2} \sum bc,\] \[\sum \frac{1}{a^3(b+c)} \ge \frac{3}{2},\] nttu wrote: Let $ a, b, c >0 , abc =1 $ . Prove that : $ \frac{1}{a^3(b+c)} + \frac{1}{b^3(a+c)} + \frac{1}{c^3(b+a)} \geq \frac{3}{2} + \frac{1}{4}[\frac{a(c-b)^2}{b+c} + \frac{b(c-a)^2}{c+a} + \frac{c(b-a)^2}{a+b}] .$ 1996,I proof the result.See《Middle school mathematics》(China wuhan)NO.11(1996). Seehere

Let $a=\frac{x^2}{yz}, b=\frac{y^2}{xz}, c=\frac{z^2}{xy}$, with $a,b,c \in \mathbb{R}^{+}$. The inequality becomes equivalent with $\sum \frac{y^3z^3}{x^3(y^3+z^3)} \ge \frac{3}{2}$. Now, let us denote $x^3y^3=p, x^3z^3=n, y^3z^3=m$, and we can easily see that the inequality is equivalent with $ \sum \frac{m}{n+p} \ge \frac{3}{2}$, which might easily be proved by using CBS ( it actually is the statement of Nessbit's Inequality )

Again, Quote: "Sorry for bringing this up again but I just recently did this problem from an inequalities packet"

From Titu's Lemma and the AM-GM inequality, $\sum_{cyc} \frac{\frac{1}{a^2}}{\frac{1}{b} + \frac{1}{c}} \ge \frac{(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})^2}{\frac{2}{a}+\frac{2}{b}+\frac{2}{c}} = \frac{\frac{1}{a} + \frac{1}{b} + \frac{1}{c}}{2} \ge \frac{3(\frac{1}{abc})^{\frac{1}{3}}}{2} = \frac{3}{2}$ QED.

There was generalization 4 here.

Here \begin{align*} \sum \frac{1}{a^3 (b + c)} &= \sum \frac{\frac{1}{a^2}}{ab + ac} \\ &\geq \frac{(\sum \frac{1}{a})^2}{2\sum ab}\text{ (by Titu)} \\ &= \frac{(\sum ab)^2}{2a^2 b^2 c^2 \sum ab} \\ &= \frac{\sum ab}{2a^2 b^2 c^2} \\ &= \frac{1}{2} \sum ab \\ &\geq \frac{3}{2}\text{ (by AM-GM)} \end{align*}as desired.

$=\sum_{cyc}\frac{b^2c^2}{a(b+c)} \geq \frac{ab+bc+ca}{2} \geq \frac{3}{2}$ by Titu's lemma and AM-GM respectively. We are thus done.

By Holder's inequality we get \[\left(\sum_{\text{cyc}} a(b+c)\right)^{\frac{1}{2}} \left(\sum_{\text{cyc}} \frac{1}{a^3(b+c)} \right)^{\frac{1}{2}} \geq \left(\frac 1a + \frac 1b + \frac 1c \right) = (ab+ac+bc)\]Thus it remains to prove that $\frac{ab+ac+bc}{2} \geq \frac{3}{2}$. This is obvious from Muirhead with $\left(\frac 43, \frac 43, \frac 13 \right) \succ (1, 1, 1)$

$a=b=c$ works Multiplying by the denominators and using $abc = 1$ gives $$\sum{a^3b^2 a^2b^3+ bc + b^4c^4} \ge 3 + \frac{3}{2}\sum{a^2b + b^2a}$$From Muirhead's Inequality we have $$\sum{a^2b + b^2a} = \sum{a^{\frac{8}{3}} b^{\frac{5}{3}} c^{\frac{2}{3}}} \le \sum{a^3b^2 + a^2b^3} $$$$\sum{a^2b + b^2a} = \sum{a^{\frac{11}{3}} b^{\frac{8}{3}} c^{\frac{5}{3}}} \le \sum{a^4b^4} $$$$ab+ac+bc \ge 3$$Adding them all together gives the desired inequality.

$$\frac {1}{a^{3}\left(b + c\right)} + \frac {1}{b^{3}\left(c + a\right)} + \frac {1}{c^{3}\left(a + b\right)}\geq \frac {3}{2}$$As $abc=1, $Multiplying $(abc)^2$ on the $LHS$, we get: $$\frac{(bc)^2}{a(b+c)} + \frac{(ca)^2}{b(c+a)}+\frac{(ab^2)}{c(a+b)} \geq \frac{(ab+bc+ca)^2}{a(b+c)+b(c+a)+c(a+b)}=\frac{(ab+bc+ca)}{2}$$[By Titu's Lemma] Again, $\frac{ab+bc+ca}{2} \geq \frac{3 \sqrt[3]{a^2b^2c^2}}{2} = \frac{3}{2}$ by AM-GM. Hence, we're done!

a variation If $a+b+c=3,$ then $$\frac{1}{a^3(b+c)}+\frac{1}{b^3(a+c)}+\frac{1}{c^3(a+b)} \geq \frac{3}{2}.$$

anduran wrote: a variation If $a+b+c=3,$ then $$\frac{1}{a^3(b+c)}+\frac{1}{b^3(a+c)}+\frac{1}{c^3(a+b)} \geq \frac{3}{2}.$$ Write$b+c = 3-a$ etc, and use Jensen on $f(x) = \frac{1}{x^3(3-x)}$.

Notice that \begin{align*} \frac {1}{a^{3}\left(b + c\right)} + \frac {1}{b^{3}\left(c + a\right)} + \frac {1}{c^{3}\left(a + b\right)} &= \frac {a^2b^2c^2}{a^{3}\left(b + c\right)} + \frac {a^2b^2c^2}{b^{3}\left(c + a\right)} + \frac {a^2b^2c^2}{c^{3}\left(a + b\right)} \\ &= \frac {b^2c^2}{ab+ac} + \frac {c^2a^2}{bc+ba} + \frac {a^2b^2}{ca+cb}. \end{align*}By Titu's Lemma, \begin{align*} \frac {b^2c^2}{ab+ac} + \frac {c^2a^2}{bc+ba} + \frac {a^2b^2}{ca+cb} \geq \dfrac{(ab+bc+ca)^2}{2(ab+bc+ca)} = \dfrac{ab+bc+ca}{2}. \end{align*}By AM-GM, note that $ab + bc + ca \geq 3\sqrt[3]{a^2b^2c^2} = 3$, as desired.

I was to lazy to read all the solutions so never mind if the solution is same. By using titu's lemma $\frac{\frac{1^2}{a^2}}{a(b+c)} + \frac{\frac{1^2}{b^2}}{b(a+c)}+\frac{\frac{1^2}{c^2}}{c(b+a)} \ge \frac{(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})^2}{2(ab+bc+ac)}$ After some simple calculations we finally get: $\frac{\frac{1^2}{a^2}}{a(b+c)} + \frac{\frac{1^2}{b^2}}{b(a+c)}+\frac{\frac{1^2}{c^2}}{c(b+a)} \ge \frac{ab+bc+ac}{2}$ Using A.M.-G.M. inequality on $ \frac{ab+bc+ac}{2}$ $\frac{ab+bc+ac}{2*3} \ge \frac{\sqrt[3]{(abc)^2)}}{2}$ hence,$\frac{\frac{1^2}{a^2}}{a(b+c)} + \frac{\frac{1^2}{b^2}}{b(a+c)}+\frac{\frac{1^2}{c^2}}{c(b+a)} \ge \frac{ab+bc+ac}{2}\ge \frac{\sqrt[3]{(abc)^2}*3}{2}\ge \frac{3}{2}$

I discussed this problem on my youtube channel (little fermat) Video in my inequalities tutorial playlist.

Substitute $a=\frac{x}{y}$, $b=\frac{y}{z}$ and $c=\frac{z}{x}$. We get, $$\sum_{cyc} \frac{y^3z}{x^3y+x^2z^2}$$ Multiply and divide by $yz$ $$\sum_{cyc} \frac{y^4z^2}{x^3y^2z+x^2yz^3} \geq \frac{(x^2y+y^2z+z^2x)^2}{2xyz(x^2y+y^2z+z^2x)} = \frac{x^2y+y^2z+z^2x}{2xyz}$$This inequality is by Cauchy-Schwarz. It is sufficient to prove that $$\frac{x^2y+y^2z+z^2x}{2xyz} \geq \frac{3}{2}$$$$\implies x^2y+y^2z+z^2x \geq 3xyz $$ This is true by AM-GM

$\sum\frac{1}{a^3(b+c)}=\sum\frac{\frac1{a^2}}{ab+ac}\geq\frac{(\frac1{a}+\frac1{b}+\frac1{c})^2}{2(ab+ac+bc)}=\frac{(ab+bc+ca)^2}{2(ab+ac+bc)}=\frac{(ab+bc+ca)}{2}\geq\frac{3\sqrt[3]{a^2b^2c^2}}{2}=\frac32$

First, let $x = \frac{1}{a}$, $y = \frac{1}{b}$, and $z = \frac{1}{c}$. Our sum is then \[\sum_{\text{cyc}} \frac{1}{\frac{1}{x^3}\left(\frac{1}{y} + \frac{1}{z} \right)} = \sum_{\text{cyc}} \frac{1}{\frac{y + z}{x^3yz}} = \sum_{\text{cyc}} \frac{1}{\frac{y + z}{x^2}} = \sum_{\text{cyc}} \frac{x^2}{y+z}\] Applying Titu and AM-GM, we have \[\sum_{\text{cyc}} \frac{x^2}{y+z} \geq \frac{(x+y+z)^2}{2(x+y+z)} = \frac{x+y+z}{2} \geq \frac{3\sqrt[3]{xyz}}{2} = \frac{3}{2}\] as desired.