Here the short version of my sol: from the first piece of inequality we get $a+b+c \geq 1$: order inequality, replacing 1 ($ab+bc+ac$ by $a+b+c$ will keep the inequality, then simplify and again do this gives $a+b+c \geq 1$ from the second part we get $a+b+c \leq 1$: $(a+b+c)(a+b+c) \leq 3(a+b+c)$ but the problem says without $(a+b+c)$ it's $\geq$ so $a+b+c \leq 1$ Therefor, a+b+c = 1 which also implies that order inequality in the second case is an equality, so a=b=c Which results in $a=b=c=\frac{1}{3}$

i don't get this solution can anybody clear it up to me?

what is it exactly you don't get? I first prove that a+b+c must be >= 1 (first part) Then I prove that a+b+c must be <= 1 at the same time, so that means a+b+c = 1 for sure. from there I call in the order inequality I used to prove the above, which states that it becomes an equality iff a=b=c. So since the sign is reversed, we must have an equality here for the given to be possible, thus a=b=c with a+b+c=1, this results in a=b=c=1/3. Is it one of the proofs of the first/second step you don't get?

actually both of them i don't understand (what is "order" ineq?)

order ineq says that 5 bills of 5 dollar + 2 bills of 1 dollar is worth more than 2 bills of 5 dollar and 5 bills of 1 dollar (so the greedy law, take most of the highes and fewest of the lowest to get best result) thus, order ineq: if a>=b>=c then a^2+b^2+c^2>=ab+bc+ca [ that is a common ineq, which can also be constructed by taking (a-b)^2+(b-c)^2+(c-a)^2>=0 ] that's quite important, a^2+b^2+c^2>=ab+bc+ca for all a,b,c in R. (so in each of the steps I could have added: assume without loss of generality that a>=b>=c) Is it clear now, or do you want me to explain it in full detail?

Adding $2(a^2+b^2+c^2)$ to both sides of the left inequality, we obtain \[2(a+b+c)^2-1 \geq 3(a^2+b^2+c^2).\] Clearly, $3(a^2+b^2+c^2) \geq (a+b+c)^2$ (easily proven by Cauchy or Power-Mean, for example and equality iff $a=b=c$), so we have \[2(a+b+c)^2-1 \geq (a+b+c)^2 \iff a+b+c \geq 1,\] since $a,b,c$ are positive. Now note that $3(a^3+b^3+c^3) \geq (a+b+c)(a^2+b^2+c^2)$ which easily follows from $(a^3+a^3+b^3)/3 \geq a^2b$, etc by AM-GM. The right side of the inequality implies that $a+b+c \leq 1$, so combining this with the above, we conclude that $a+b+c=1$, so equality must hold, or $a=b=c=1/3$.

Yes. ThAzN1 wrote: ... so equality must hold, or $a=b=c=1/3$. => here for the case it may be unclear: you substitute the given of equalness in the second step, to find that (like I said) to find another equality, from which you can conclude like I said that a=b=c.

thanks, i get it now. I think order ineq is the ineq known in my country as "rearrangament ineq": given $a_{1} \geq \cdots \geq a_{n}$ and $b_{1} \geq \cdots \geq b_{n}$ we have: $a_{1} \cdot b_{1} + \cdots + a_{n} \cdot b_{n} \geq a_{1} \cdot b_{ \tau (1) } \cdots a_{n} \cdot b_{ \tau (n) } \geq a_{1} \cdot b_{n} + \cdots + a_{n} \cdot b_{1}$ for any permutation $\tau$ of $(b_{1}, \cdots, b_{n})$.

yes, that is another (even better) name for it, rearrangement

$\frac{4(a+b+c)^2}{3}\geq4(ab+ac+bc)\geq a^2+b^2+c^2+1\geq \frac{(a+b+c)^2}{3}+1 \Rightarrow a+b+c\geq1$ $a^2+b^2+c^2 \geq 3(a^3+b^3+c^3) \geq (a+b+c)(a^2+b^2+c^2) \Rightarrow a+b+c\leq1$ Then $ a=b=c=\frac{1}{3}$

Mamat wrote: $\frac{4(a+b+c)^2}{3}\geq4(ab+ac+bc)\geq a^2+b^2+c^2+1\geq \frac{(a+b+c)^2}{3}+1 \Rightarrow a+b+c\geq1$ $a^2+b^2+c^2 \geq 3(a^3+b^3+c^3) \geq (a+b+c)(a^2+b^2+c^2) \Rightarrow a+b+c\leq1$ Then $ a=b=c=\frac{1}{3}$ Nice! I did the same thing, after spending a ton of time on it. A bit of an explanation: The first one follows from a^2+b^2+c^2+2ab+2ac+2bc >= 3ab+3ac+3bc, which is true by a^2+b^2-2ab >=0, so adding all cyclic permutations of it and dividing by two, gives a^2+b^2+c^2-ab-ac-bc >= 0. (a^2+b^2+c^2)(1+1+1) >= (a+b+c)^2 by C-S, so the third-fourth part is true. thanks @below for clarifying a final part, I thought you would have to use some inequality but I wasn’t considering just expanding

Claim: $3(a^3+b^3+c^3)\ge (a+b+c)(a^2+b^2+c^2)$ Proof: Rearrange the inequality to \[2a^3+2b^3+2c^3\ge a^2b+ab^2+b^2c+bc^2+c^2a+ca^2.\]It suffices to prove \[a^3+b^3\ge a^2b+ab^2\]but this is equivalent to \[(a-b)^2(a+b)\ge 0\]after rearranging, clearly true since $a$, $b$, and $c$ are positive reals. $\blacksquare$