April wrote: Let $ a,\, b,\, c$ be side lengths of a triangle and $ a+b+c = 3.$ Find the minimum of \[ a^{2}+b^{2}+c^{2}+\frac{4abc}{3}\] P.S. Thank you, Zhaoli. Since $ a,b,c$ are sides of a triangle we can change $ a=x+y$, $ b=y+z$ and $ c=z+x$ with $ x+y+z=\frac{3}{2}$ So we have: $ a^{2}+b^{2}+c^{2}+\frac{4abc}{3}=\frac{(a^{2}+b^{2}+c^{2})(a+b+c)+4abc}{3}=$ $ \frac{2((x+y)^{2}+(y+z)^{2}+(z+x)^{2})(x+y+z)+4(x+y)(y+z)(z+x)}{3}=$ $ \frac{4(x^{3}+y^{3}+z^{3}+3x^{2}y+3xy^{2}+3y^{2}z+3yz^{2}+3z^{2}x+3zx^{2}+5xyz)}{3}=$ $ \frac{4((x+y+z)^{3}-xyz)}{3}=\frac{4(\frac{26}{27}(x+y+z)^{3}+(\frac{x+y+z}{3})^{3}-xyz)}{3}\geq$ $ \geq \frac{4(\frac{26}{27}(x+y+z)^{3}}{3}=\frac{13}{3}$

the third degree Schur's inequality $ (a+b+c)^{3}+9abc\geq\ 4(a+b+c)(ab+ac+bc)---\to abc\geq \frac{1}{9}(12(ab+ac+bc)-27)$ and $ \frac{1}{3}(a+b+c)^{2}\geq\ ab+ac+bc$ from this $ a^{2}+b^{2}+c^{2}+\frac{4abc}{3}=(a+b+c)^{2}-2ab-2ac-2bc+\frac{4}{3}*\frac{1}{9}(12(ab+ac+bc)-27)=3^{2}-2(ab+ac+bc)+\frac{16}{9}(ab+ac+bc)-4=5-\frac{2}{9}(ab+ac+bc)\geq\ 5-\frac{6}{9}=\frac{13}{4}$ a=b=c=1

Let x=$ \frac{a}{3}$ , y=$ \frac{b}{3}$ , z=$ \frac{c}{3}$ so that $ x+y+z =1.$ Now $ a^{2}+b^{2}+c^{2}$ +$ \frac{4ab}{3}$ =$ (a+b+c)^{2}$- 2$ \sum$ab +$ \frac{4ab}{3}$ which after substitution for a,b c, in terms of x,y,z and using a+b+c =3 becomes $ 9-18(xy+yz+zx-2xyz)$. We will find the maximum of $ xy+yz+zx-2xyz.$ The identity $ x^{3}+y^{3}+z^{3}-3xyz = (x+y+z)(x^{2}+y^{2}+z^{2}-xy-yz-zx)$ , after replacing x+y+z by 1 and some re-arrangement gives $ xy+yz+zx-2xyz$ =$ x^{2}(1-x)$ + $ y^{2}(1-y)$ +$ z^{2}(1-z)$ +$ xyz$ Assume that x $ \leq$ y $ \leq$ z . Then 1-x $ \geq$ 1-y $ \geq$1-z Writing $ \sum$$ x^{2}(1-x)$ as $ \sum(x.x.(1-x) )$ ,Tchebychefs inequality gives $ \frac{\sum(x.x.(1-x)}_{3}$ $ \leq$ $ \frac{\sum(x)}_{3}$.$ \frac{\sum(x)}_{3}$.$ \frac{\sum(1-x)}_{3}$ which gives $ \sum{(x^{2}.(1-x)}$ $ \leq$ $ \frac{6}{27}$ with equality iff x=y=z, Finally by arithmetic-geometric inequality xyz $ \leq$ $ (x+y+z)^{3}/27$ =$ \frac{1}{27}$ with equality iff x=y=z. Putting the two inequalities together gives $ \frac{\sum(x.x.(1-x)}_{3}$ + $ xyz$ $ \leq$ $ \frac{7}{27}$ with equality iff x=y=z. Thus the greatest value of $ xy+yz+zx-2xyz$ is $ \frac{7}{27}$ and it follows that the least value of $ 9-18(xy-yz-zx-2xyz)$ is $ \frac{13}{3}$ This problem is essentially Q1 from IMO 1984.

From substituting a=x+y,b=y+z and c=z+x, we get, that x+y+z=3/2. After changing $ a^{2}+b^{2}+c^{2}+4/3abc$ by x,y and z, we get: $ a^{2}+b^{2}+c^{2}+4abc/3 = 1/6(27-8xyz)\geq 1/6(27-1) = 13/3$,(By Cauchy) with equality if x=y=z=3/2,or a=b=c=1. So min(a^{2}+b^{2}+c^{2}+4abc/3)=13/3.

I think it is true for all $a,b,c$ such that $a+b+c=3$ . They need not be sides of a triangle. My solution ---- Let $f(a,b,c)=a^2+b^2+c^2+\frac{4abc}{3}$ Lemma 1-- At least two of $a,b,c$ are less than $\frac {3}{2}$ Proof-- Suppose not . We get contradiction. Lemma 2-- Suppose $a$ is less than $\frac {3}{2}$ then $f(a,b,c) \geq f(a,\frac {b+c}{2},\frac {b+c}{2})$ Proof-- Equivalent to $(\frac{1}{2}-\frac{a}{3})(b-c)^2 \geq 0$ Now using Lemma 1 we get $\frac{b+c}{2}$ and $a$ less than $\frac {3}{2}$ Thus we have minimum at $a=b=c$ Thus we get the answer $\frac{13}{3}$

zaya_yc wrote: 5-\frac{6}{9}=\frac{13}{4}$ a=b=c=1 Looks like no one noticed the typo!

LOL I think they added triangle sides to keep up the computational structure of the exam

April wrote: Let $ a,\, b,\, c$ be side lengths of a triangle and $ a+b+c = 3$. Find the minimum of \[ a^{2}+b^{2}+c^{2}+\frac{4abc}{3}\] Let $a,b,c$ be positive numbers such that $a+b+c=3.$ Prove that$$a^{2}+b^{2}+c^{2}+\frac{4}{3}abc\ge \frac{13}{3} .$$Let $a,b,c$ be positive numbers such that $a+b+c=1.$ Prove that$$ a^2+b^2+c^2+4abc \ge \frac{13}{27} .$$Let $a,b,c$ be side lengths of a triangle with $a+b+c = 1$. Prove that $$a^2 +b^2 +c^2 +4abc <\frac12$$1990 ITAMO Let $a,b,c$ be side lengths of a triangle with $a+b+c=2.$ Prove that \[\frac{52}{27}\le a^{2}+b^{2}+c^{2}+2abc<2\]Let $a,b,c$ be side lengths of a triangle with $a+b+c = 1.$ Prove that

China North MO 2006: Let$a,b,c$ are positive numbers such that $a+b+c=3$, show that:\[\frac{a^{2}+9}{2a^{2}+(b+c)^{2}}+\frac{b^{2}+9}{2b^{2}+(c+a)^{2}}+\frac{c^{2}+9}{2c^{2}+(a+b)^{2}}\leq 5\]