It can be rewritten $(a+b)^2 + c^2 \geq 2(a+b)c$ which is just AM >= GM with equality iff $a+b = c$

umm... yes. Some people call it $(a+b-c)^2$ as well.

should just be $\mathbb{R}^+$

Since $a,b,c >0$, multiplying throughout by $abc$ we have $a^2+b^2+c^2 \geq 2bc+2ac-2ab \iff [(a+b)-c]^2 \geq 0$, trivially true. Equality occurs iff $(a+b)-c=0$, i.e. $a+b=c$.

$(a+b–c) ^2=0$ for equality case

Peter wrote: Prove that for all $a,b,c \in \mathbb{R}^+_0$ we have \[\frac{a}{bc}+\frac{b}{ac}+\frac{c}{ab} \ge \frac2a+\frac2b-\frac2c\]and determine when equality occurs. Proof of Zhangyunhua:By AM-GM,$$\frac{a}{bc}+\frac{b}{ac}+\frac{c}{ab} =\frac{(a+b)^2}{abc}+\frac{c}{ab}-\frac{2}{c}\ge\frac{2(a+b)}{ab}-\frac{2}{c}= \frac2a+\frac2b-\frac2c.$$

can we use cauchy schwarz inequality here?

By AM-GM, $(a+b)^2+c^2 \geq 2ac+2bc$. Simplifying, we get $a^2+b^2+c^2 \geq 2bc+2ac-2ab$, and dividing by $abc$ yields the desired inequality.

$(a+b-c)^2\Rightarrow a^2+b^2+c^2\ge2bc+2ac-2ab\Rightarrow\frac a{bc}+\frac b{ac}+\frac c{ab}\ge\frac2a+\frac2b-\frac2c$

Peter wrote: Prove that for all $a,b,c \in \mathbb{R}^+_0$ we have \[\frac{a}{bc}+\frac{b}{ac}+\frac{c}{ab} \ge \frac2a+\frac2b-\frac2c\]and determine when equality occurs. $/ frac {a} {bc} + / frac {b} {ac} + / frac {c} {ba} = / frac {a ^ 2 + b ^ 2 + c ^ 2} {abc} and $/ frac {2} {a} + / frac {2} {b} + / frac {2} {c} = / frac {2 (bc + ac-ab)} {abc}$ then: $/ frac {a ^ 2 + b ^ 2 + c ^ 2} {abc}$ is greater than $/ frac {2 (bc + ac-ab)} {abc}$ and this is true if and only if $a ^ 2 + b ^ 2 + c ^ 2$ is greater than or equal to $2 (bc + ac-ab)$ which is true if and only if $(a + b - c) ^ 2$ is greater than or equal to 0 which is true.

e61442289 wrote: Peter wrote: Prove that for all $a,b,c \in \mathbb{R}^+_0$ we have \[\frac{a}{bc}+\frac{b}{ac}+\frac{c}{ab} \ge \frac2a+\frac2b-\frac2c\]and determine when equality occurs. $\frac {a} {bc} + \frac {b} {ac} + \frac {c} {ba} = \frac {a ^ 2 + b ^ 2 + c ^ 2} {abc}$ and $\frac {2} {a} + \frac {2} {b} + \frac {2} {c} = \frac {2 (bc + ac-ab)} {abc}$ then: $\frac {a ^ 2 + b ^ 2 + c ^ 2} {abc}$ is greater than $\frac {2 (bc + ac-ab)} {abc}$ and this is true if and only if $a ^ 2 + b ^ 2 + c ^ 2$ is greater than or equal to $2 (bc + ac-ab)$ which is true if and only if $(a + b - c) ^ 2$ is greater than or equal to $0$ which is true. FTFY. For the fraction $\LaTeX$, it starts with "\" not "/". Edit: What does $\mathbb{R}^+_0$ mean in the question? Sorry I am not familiar with the notation.

We have $\frac{a}{bc}+\frac{b}{ac}+\frac{c}{ab} = \frac{a^2+b^2+c^2}{abc}$ and $\frac2a+\frac2b-\frac2c = \frac{2bc + 2ac - 2ab}{abc}$. They both have the same denominators so we need just prove that $a^2+b^2+c^2\geq2bc+2ac-2ab$. Note that $a^2+b^2+c^2 - 2bc - 2ac + 2ab = (a + b - c)^2\geq0$, showing the given inequality. Equality occurs when $a + b = c$.

Since $a,b,c>0$ we multiply all by abc So we are left with: $a^2+b^2+c^2\ge 2bc+2ac-2ab$ $\rightarrow (a+b)^2+c^2\ge 2c(a+b)$ which is true for $AM-GM$

Peter wrote: Prove that for all $a,b,c \in \mathbb{R}^+_0$ we have \[\frac{a}{bc}+\frac{b}{ac}+\frac{c}{ab} \ge \frac2a+\frac2b-\frac2c\]and determine when equality occurs.

e61442289 wrote: Peter wrote: Prove that for all $a,b,c \in \mathbb{R}^+_0$ we have \[\frac{a}{bc}+\frac{b}{ac}+\frac{c}{ab} \ge \frac2a+\frac2b-\frac2c\]and determine when equality occurs. $/ frac {a} {bc} + / frac {b} {ac} + / frac {c} {ba} = / frac {a ^ 2 + b ^ 2 + c ^ 2} {abc} and $/ frac {2} {a} + / frac {2} {b} + / frac {2} {c} = / frac {2 (bc + ac-ab)} {abc}$ then: $/ frac {a ^ 2 + b ^ 2 + c ^ 2} {abc}$ is greater than $/ frac {2 (bc + ac-ab)} {abc}$ and this is true if and only if $a ^ 2 + b ^ 2 + c ^ 2$ is greater than or equal to $2 (bc + ac-ab)$ which is true if and only if $(a + b - c) ^ 2$ is greater than or equal to 0 which is true. LATEX

e61442289 wrote: Peter wrote: Prove that for all $a,b,c \in \mathbb{R}^+_0$ we have \[\frac{a}{bc}+\frac{b}{ac}+\frac{c}{ab} \ge \frac2a+\frac2b-\frac2c\]and determine when equality occurs. $\frac{a}{bc} + \frac{b}{ac} + \frac{c}{ba} = \frac{a^2 + b^2 + c^2}{abc}$ and $\frac{2}{a}+ \frac{2}{b} + \frac{2}{c} = \frac{2 (bc + ac-ab)}{abc}$ then: $\frac {a ^ 2 + b ^ 2 + c ^ 2}{abc}$ is greater than $\frac{2 (bc + ac-ab)}{abc}$ and this is true if and only if $a^2 + b^2 + c^2$ is greater than or equal to $2 (bc + ac-ab)$ which is true if and only if $(a + b - c) ^ 2$ is greater than or equal to 0 which is true. I fixed the LaTeX but idk about this proof

I will present mathmax12's solution (which he shared with me and gave me permission to show.): Multiplying by $abc$ gives you $a^2+b^2+c^2 \ge 2bc+2ac-2ab$, so $((a+b)-c)^2 \ge 0$, which is always true by the trivial inequality, equality holds when $\boxed{a+b=c}.$