$ \Leftrightarrow \frac {1}{4} < \frac {(a + b)(a + c)(b + c)}{(a + b + c)^3}\le \frac {8}{27}$ 1). $ \sqrt [3]{(a + b)(a + c)(b + c)}\le \frac {(a + b) + (a + c) + (b + c)}{3} = \frac {2.(a + b + c)}{3}\Rightarrow$ $ \Rightarrow (a + b)(a + c)(b + c)\le \frac {8}{27}.{(a + b + c)^3}\Rightarrow \frac {(a + b)(a + c)(b + c)}{(a + b + c)^3}\le \frac {8}{27}.$ 2). If $ a = y + z; \ b = x + z; \ c = x + y,$ then: $ a + b = x + y + 2z; \ a + c = x + 2y + z; \ b + c = 2x + y + z; \ a + b + c = 2.(x + y + z)$ and $ \Leftrightarrow \frac {1}{4} < \frac {(a + b)(a + c)(b + c)}{(a + b + c)^3}\Leftrightarrow$ $ \Leftrightarrow 2.(x + y + z)^3 < (2x + y + z)(x + 2y + z)(x + y + 2z);\ (\forall)x,y,z > 0.$ 3).$ (2x + y + z)(x + 2y + z)(x + y + 2z)=[(x+y+z)+x][(x+y+z)+y][(x+y+z)+z]=$ $ =[(x+y+z)^2+(x+y+z)(x+y)+xy][(x+y+z)+z]=$ $ =(x+y+z)^3+z(x+y+z)^2+(x+y)(x+y+z)^2+z(x+y+z)(x+y)+$ $ +xy(x+y+z)+xyz=2.(x+y+z)^3+(x+y+z).(xy+xz+yz)+xyz>2.(x+y+z)^3.$

We have $ AI^2 = (p - a)^2 + r^2 = (p - a)^2 + \frac {S^2}{p^2} = (p - a)^2 + \frac {(p - a)(p - b)(p - c)}{p} = \frac {bc(p - a)}{p}$ $ AD^2 = \frac {4bc\cdot p(p - a)}{(b + c)^2}$ Therefore $ \frac {AI}{AD} = \sqrt {\frac {\frac {bc(p - a)}{p}}{\frac {4bc\cdot p(p - a)}{(b + c)^2}}} = \dfrac{b + c}{2p} = \dfrac{b + c}{a + b + c}$. So we need to prove that $ \frac {1}{4} < \frac {(a + b)(a + c)(b + c)}{(a + b + c)^3}\le \frac {8}{27}$ For the right part see mihai miculita's proof. For left part note that $ (a + b + c)^3 - 3(a + b)(b + c)(c + a) = a^3 + b^3 + c^3$. So we need to prove that $ (a + b)(b + c)(c + a) > a^3 + b^3 + c^3$. But $ (a + b)(b + c)(c + a) > a(a + b)(c + a)= a(a^2 + bc + a(b + c)) > a^2(a + b + c) > 3a^3$ Summing analogies we get the result.

It appears in Engel's Problem solving strategies in the inequalities section as an IMO problem. Let CD=p, then $ \frac{AI}{ID}=\frac{b}{p}=\frac{b+c}{a}$, so $ \frac{AI}{AD}=\frac{b+c}{a+b+c}$ and we have the inequality $ \frac {1}{4} < \frac {(a + b)(a + c)(b + c)}{(a + b + c)^3}\le \frac {8}{27}$ The right side is clear by AM-GM. For the left side, by the triangle inequality: $ (a + b - c)(a - b + c)( - a + b + c) > 0$ Let $ u = a + b + c$, $ v = ab + bc + ac$, $ w = abc$, this is equivalent to $ - u^3 + 4uv - 8w > 0$ However, the left side is equivalent to $ - u^3 + 4uv - 4w > 0$ so we are done.

Approach by e.lopes: we know that $ \frac {AI}{AA'} = \frac {b + c}{a + b + c}$ so, using the analogous equalities, we have to prove that $ \frac {1}{4} < \frac {(a + b)(a + c)(b + c)}{(a + b + c)^3} \leq \frac {8}{27}$. right side: by $ AM - GM$, we have: $ (\frac {(\frac {a + b}{2}) + (\frac {a + c}{2}) + (\frac {c + b}{2})}{3})^3 \ge (\frac {a + b}{2})(\frac {c + b}{2})(\frac {a + c}{2})$ (easy to see that this is exactly the right side) left side: let $ a = x + y$, $ b = y + z$ and $ c = x + z$ and $ s = x + y + z$. we have to prove that $ 2s^3 < (s + x)(s + y)(s + z)$ but $ (s + x)(s + y)(s + z) = s^3 + s^2(x + y + z) + s(xy + xz + zx) + xyz$ $ = 2s^3 + + s(xy + yz + zx) + xyz$. E.L

Note in (a) $I$ doesn't need to be the incentre of $ABC$. Just consider areal coordinates, where $A=(1,0,0)$, $B=(0,1,0)$, $C=(0,0,1)$ and $I=(p,q,r)$. We have $\frac{AI}{AA'}=\frac{AA'-IA'}{AA'}=1-\frac{IA'}{AA'}=1-p=q+r$. Similarly we have $\frac{BI}{BB'}=p+r$ and $\frac{CI}{CC'}=p+q$, therefore $\frac{AI\cdot BI\cdot CI}{AA'\cdot BI'\cdot CI'}=(p+q)(p+r)(q+r)$. By the AM-GM inequality $\frac{AI\cdot BI\cdot CI}{AA'\cdot BI'\cdot CI'}=(p+q)(p+r)(q+r)\le\left(\frac{(p+q)+(p+r)+(q+r)}{3}\right)^3=\frac{8}{27}$. For part (b), let $a=BC$, $b=AC$ and $c=AB$. Since $I$ is the incentre of $ABC$, $p=\frac{a}{a+b+c}$, $q=\frac{b}{a+b+c}$ and $r=\frac{c}{a+b+c}$. Therefore we have $\frac{AI\cdot BI\cdot CI}{AA'\cdot BI'\cdot CI'}=\frac{(a+b)(a+c)(b+c)}{(a+b+c)^3}$. Now put $a=x+y$, $b=x+z$ and $c=y+z$. We want to prove $\frac{(2x+y+z)(x+2y+z)(x+y+2z)}{8(x+y+z)^3}>\frac{1}{4}$. With some brute-force computation we easily obtain an obviously true inequality.

Joao Pedro Santos wrote: Now put $a=x+y$, $b=x+z$ and $c=y+z$. We want to prove $\frac{(2x+y+z)(x+2y+z)(x+y+2z)}{8(x+y+z)^3}>\frac{1}{4}$. With some brute-force computation we easily obtain an obviously true inequality. In IMO compendium they also used Raavi supstitution, and got the same inequality like you, but use supstitution $T=x+y+z$ and your computation will get more easier. Sorry for reviving this post, but this week I was working on bisector theorem and this was an example in my book. And I found interesting solution for left inequality: $\frac{(b+c)(c+a)(a+b)}{(a+b+c)^{3}}>\frac{1}{4}$ $\frac{b+c+a-a}{a+b+c}\cdot \frac{c+a+b-b}{a+b+c}\cdot \frac{a+b+c-c}{a+b+c}>\frac{1}{4}$ $\prod \left(1-\frac{a}{a+b+c}\right)>\frac{1}{4}$. Take $x=\frac{a}{a+b+c}, y=\frac{b}{a+b+c}, z=\frac{c}{a+b+c}$ so $x+y+z=1$ and our inequality is equivalent to: $(1-x)(1-y)(1-z)=1-(x+y+z)+xy+xz+yz-xyz=\frac{ab+ca+bc}{4s^2}-\frac{abc}{8s^3}>\frac{1}{4}$ Using well-known identities $ab+ca+bc=r^2+s^2+4Rr$ and $abc=4Rrs$ and multiplying it with $4s^2>0$ we get $r(2R+r)>0$ q.e.d.

SQ Indonesia TST 2006: Let $x,y,z>0,x+y+z=3.$ Prove that $$\sqrt {\frac{2}{ \frac{1}{x^3}+1}}+\sqrt {\frac{2}{ \frac{1}{y^3}+1}}+\sqrt {\frac{2}{ \frac{1}{z^3}+1}}\leq 3.$$Solution: $$\sum_{cyc}\sqrt{\frac{2}{\frac{1}{x^3}+1}}\leq \sum_{cyc}\sqrt{\frac{2}{2\sqrt{\frac{1}{x^3}}}}=\sum_{cyc}\sqrt[4]{x^3}\leq\sum_{cyc}\frac{3x+1}{4}=3.$$

2) We have to prove that $ \frac{(a+b)(b+c)(a+c)}{(a+b+c)^3}> \frac{1}{4} $ or $ \frac{1}{8}(1+\frac{a+b-c}{a+b+c})(1+\frac{b+c-a}{a+b+c})(1+\frac{c+a-b}{a+b+c})> \frac{1}{8}(1+\frac{a+b-c+b+c-a+c+a-b}{a+b+c})=\frac{1}{4} $ ( By Bernoulli)

The right ineqality is correct for every point inside ABC, eqality hilds iff I is the center of gravity. It is equivalent to П(a+b)/(a+b+c)^3<=8/27, where a is the area of the triangle BIC, b is the area of the triangle AIC, c the area of the triangle AIB. Obvious by AM-GM.

Here's an approach which may seem longer than the other solutions in this thread but which makes the inequality part very easy. As others have stated, the inequality reduces to proving \[\dfrac14 < \dfrac{(a+b)(b+c)(c+a)}{(a+b+c)^3}\leq\dfrac{8}{27}.\]Let $a=x+y$, $b=y+z$, $c=z+x$ for $x,y,z>0$ (Ravi Substitution). Then the expression in the middle becomes \[\dfrac{(2x+y+z)(x+2y+z)(x+y+2z)}{8(x+y+z)^3},\]and so the inequality becomes \begin{align*}2&<\dfrac{(2x+y+z)(x+2y+z)(x+y+2z)}{(x+y+z)^3}\leq\dfrac{64}{27}\\\iff 2&<\left(1+\dfrac{x}{x+y+z}\right)\left(1+\dfrac{y}{x+y+z}\right)\left(1+\dfrac{z}{x+y+z}\right) \leq\dfrac{64}{27}.\end{align*}Now let $p=\tfrac{x}{x+y+z}$, $q=\tfrac{y}{x+y+z}$, $r = \tfrac{z}{x+y+z}$, so that $p+q+r=1$. Then the inequality becomes \[2<(1+p)(1+q)(1+r)\leq\dfrac{64}{27},\]which is much easier to work with. Left Hand Side: Note that \[(1+p)(1+q)(1+r) = 2 + (pq+qr+rp+pqr) > 2.\] Right Hand Side: Note that \[(1+p)(1+q)(1+r)\leq\left(\dfrac{3+p+q+r}3\right)^3 = \dfrac{64}{27}.\] We are done.

In triangle,easy prove this: \[{\frac { \left( a+b \right) \left( b+c \right) \left( c+a \right) }{ \left( a+b+c \right) ^{3}}}\geq \frac{1}{4}+{\frac {5}{36}}\,{\frac {\sqrt {3}r}{ s}}\] But hard is this: \[{\frac { \left( m_a+m_b \right) \left( m_b+m_c \right) \left( m_c+m_a \right) }{ \left( m_a+m_b+m_c \right) ^{3}}}\geq \frac{1}{4}+{\frac {5}{36}}\,{\frac {\sqrt {3}r}{ s}}\]

Is anyone here that TA from Awesome Math camp who in class today said he posted a solution to this problem? Edit: Sorry if this is off topic.

Possibly djmathman

It was djmathman. I should've asked for his autograph.

Anyone tried Euler-Gergonne Theorem here? It trivializes the right side by a whole different level. The left side is just trig bash.

First Note that $\frac{AI}{AA'} = \frac{\frac{bc.\cos{\frac{A}{2}}}{p}}{\frac{2bc.\cos{\frac{A}{2}}}{b+c}} = \frac{b+c}{a+b+c}$ so we need to prove $ \frac {1}{4} < \frac {(a + b)(a + c)(b + c)}{(a + b + c)^3}\le \frac {8}{27}$. Right Side : $2 = \frac{b+c}{a+b+c} + \frac{c+a}{a+b+c} + \frac{a+b}{a+b+c} \ge 3\sqrt [3]{\frac{(a + b)(a + c)(b + c)}{(a + b + c)^3}} \implies \frac{8}{27} \ge \frac {(a + b)(a + c)(b + c)}{(a + b + c)^3}$ Left Side : From Ravi Transformation we have there exists $x,y,z$ such that $a = x+y, b = y+z, c = z+x, s = x+y+z$. we have to prove $ 2s^3 < (s + y)(s + z)(s + x) = s^3 + s^2(x+y+z) + s(xy+yz+zx) + xyz = 2s^3 + s(xy+yz+zx) + xyz$ which is True.

Let $a,b,c\geq 0 $ and $a+b+c=1.$ Prove that $$(a+1)(2b+1)(c+1)\leq \frac{343}{108}$$$$(a+1)(b^2+1)(c+1)\leq \frac{9}{4}$$$$(a+1)(2\sqrt b+1)(c+1)\leq \frac{11979}{3125}$$11#

Here are my proofs for $(a+b+c)^3<4(a+b)(b+c)(c+a)$ without using Ravi Transformation . Proof 1 $(a+b+c)^3<4(a+b)(b+c)(c+a)$ $\Longleftrightarrow a^3+b^3+c^3+3(a+b)(b+c)(c+a)<4(a+b)(b+c)(c+a)$ $\Longleftrightarrow a^3+b^3+c^3 <(a+b)(b+c)(c+a)$ $\Longleftrightarrow a^3 - (b+c)a^2 - (b+c)^2 a +(b+c)(b-c)^2 - 4abc < 0$ $\Longleftrightarrow - (a+b-c)(b+c-a)(c+a-b) -4abc < 0\ \because a+ b > c,\ b+c>a, \ c+a>b.$ Proof 2 Let $a+b+c = 1$, W.L.O.G. By Heron, $4S : = 4[ABC] = \sqrt {(1-2a)(1-2b)(1-2c)} \Longleftrightarrow (1-2a)(1-2b)(1-2c) = 16S^2.$ $\therefore 4(a+b)(b+c)(c+a) - (a+b+c)^3$ $= 4(1-a)(1-b)(1-c) - 1$ $= 4(ab+bc+ca) - 4abc - 1$ $= (1-2a)(1-2b)(1-2c) + 4abc$ $= 16S^2 + 4abc > 0.$

My proof 3 for $(a+b+c)^3<4(a+b)(b+c)(c+a).$ $$4(a+b)(b+c)(c+a) - (a+b+c)^3= (b+c-a)a^2+(c+a-b)b^2+(a+b-c)c^2+2abc > 0.$$

Note that using the angle bisector theorem a couple times, $\frac{AI}{AA'} = \frac{b+c}{a+b+c}$. From here the inequality becomes $$\frac 14 < \frac{(a+b)(b+c)(c+a)}{(a+b+c)^3} \le \frac 8{27}.$$The right side is true by AM-GM. For the left side, let $a = x+y, b = y+z, c = z+x$ with $x+y+z = s$. Then the inequality becomes $$2s^3 < (s+x)(s+y)(s+z)$$which is true as $$(s+x)(s+y)(s+z) = s^3 + (x+y+z)s + (xy+yz+zx)s + xyz = 2s^3 + (xy+yz+zx)s + xyz > 2s^3. $$$\square$