# Let $a,b,c \in [0,1[$ such that $a+b+c=2$. Prove that $a^3+b^3+c^3+2abc\leq 2$

Let $$a,b,c \in [0,1[$$ such that $$a+b+c=2$$. Prove that $$a^3+b^3+c^3+2abc\leq 2$$

My attempt:

put $$a=x^{\frac{2}{3}}$$,$$b=y^{\frac{2}{3}}$$,$$c=z^{\frac{2}{3}}$$

\begin{align*} &x^{2}+y^{2}+z^{2}+2x^{\frac{2}{3}}y^{\frac{2}{3}}z^{\frac{2}{3}} \\ &\quad \leq x^2+y^2+z^2+\frac{2}{3}(x^{2}+y^{2}+z^{2})\\ &\quad =\frac{5}{3}(x^{2}+y^{2}+z^{2}), \end{align*}

because $$\frac{a+b+c}{3}\geq (abc)^{\frac{1}{3}}$$. And

\begin{align*} \frac{5}{3}(x^{2}+y^{2}+z^{2}) &= \frac{5}{3}(x^{\frac{1}{3}} \cdot x^{\frac{5}{3}} + y^{\frac{1}{3}} \cdot y^{\frac{5}{3}} + z^{\frac{1}{3}} \cdot z^{\frac{5}{3}}) \\ &\leq \frac{5}{3} \sqrt{x^{\frac{10}{3}} + y^{\frac{10}{3}} + z^{\frac{10}{3}}}\sqrt{x^{\frac{2}{3}} + y^{\frac{2}{3}} + z^{\frac{2}{3}}}\\ &= \frac{5\sqrt{2}}{3}\sqrt{x^{\frac{10}{3}} + y^{\frac{10}{3}} + z^{\frac{10}{3}}} \end{align*}

Now I need to find a good frame to $$\sqrt{x^{\frac{10}{3}}+y^{\frac{10}{3}}+z^{\frac{10}{3}}}$$, or $$\sqrt{a^5+b^5+c^5}$$ for make

$$\frac{5\sqrt{2}}{3}\sqrt{x^{\frac{10}{3}}+y^{\frac{10}{3}}+z^{\frac{10}{3}}}\leq 2.$$

I have two question:

1. Is my attempt correct? (I mean if my algebraic manipulation is true.)

2. Can you find a good frame to $$\sqrt{x^{\frac{10}{3}}+y^{\frac{10}{3}}+z^{\frac{10}{3}}}$$, or $$\sqrt{a^5+b^5+c^5}$$ for make $$\frac{5\sqrt{2}}{3}\sqrt{x^{\frac{10}{3}}+y^{\frac{10}{3}}+z^{\frac{10}{3}}}\leq 2 \quad ?$$

If you have an other method, you can post it, but it will be nice if you can complete my attempt.

• If $a$ and $b$ are close to $1$ (so that $c=2-a-b$ is close to $0$), then $\frac{5}{3}(x^2+y^2+z^2)$ is close to $\frac{10}{3}$ (which is of course larger than $2$). So, there is no hope that your approach will prove the proposed inequality. To see why this happens, note that the original inequality becomes an equality as one of $a$, $b$, or $c$ goes to $0$. On the other hand, the AM–GM inequality will cause the quantity to increase a lot when $a, b, c$ differ a lot. Commented Jun 9, 2022 at 0:53
• Also, you must make use of the condition $a, b, c < 1$ somewhere in your solution, for otherwise the inequality is false only with the constraints $a,b,c\geq 0$ and $a+b+c=2$. Commented Jun 9, 2022 at 2:13
• Following on @SangchulLee's comment, is the right-side bracket intended to be inverted? If it's not a typo, what is it intended to mean? Commented Jun 9, 2022 at 3:26
• @EricSnyder: the reversed square bracket denotes the open end of an interval in many countries. So $[0,1[$ means the same as $[0,1)$. Commented Jun 9, 2022 at 10:14
• @TonyK Thanks for the heads-up. I did not know about that convention. Commented Jun 9, 2022 at 10:15

New proof:

Using $$0 \le (1 - a)(1 - b) = 1 - a - b + ab$$, we have $$ab \ge a + b - 1$$.

We have \begin{align*} &a^3+b^3+c^3 + 2abc\\ \le\,& a^2 + b^2 + c^2 + 2abc\\ =\,& (a + b)^2 - 2ab + c^2 + 2abc\\ =\,& (a + b)^2 - 2ab(1 - c) + c^2\\ \le\,& (a + b)^2 - 2(a + b - 1)(1 - c) + c^2 \\ =\,& (2 - c)^2 - 2(1-c)(1-c) + c^2 \\ =\,& (2 - c)^2 - (1-c)^2 + c^2 - (1-c)^2\\ =\,& 1 \cdot (3-2c) + (2c-1)\cdot 1\\ =\,& 2. \end{align*} We are done.

Old proof:

WLOG, assume that $$c = \min(a,b,c)$$.

Using $$0 \le (1 - a)(1 - b) = 1 - a - b + ab$$, we have $$ab \ge a + b - 1$$.

We have \begin{align*} &a^3 + b^3 + c^3 + 2abc\\ =\,& (a + b)^3 - 3ab(a + b) + c^3 + 2abc \\ =\,& (a + b)^3 - ab(3a + 3b - 2c) + c^3\\ \le\,& (a + b)^3 - (a + b - 1)(3a + 3b - 2c) + c^3\\ =\,& c^2 - c + 2\\ \le\,& 2. \end{align*}

We are done.

• Note to anyone confirming this: it is correct, but the second but last step is a nuisance due to all the signs and cancellations, I got $8-12c+6c^2-c^3-6+11c-5c^2+c^3$ out of that expression. Is there an easier way to do it? Commented Jun 9, 2022 at 10:59
• @SuzuHirose I added another proof. Commented Jun 9, 2022 at 11:41
• @River Li nice proofs,thank you
– user1065479
Commented Jun 9, 2022 at 11:45
• @anaas You are welcome. Commented Jun 9, 2022 at 11:46
• @anaas The Cauchy-Schwarz Master Class? Commented Jun 9, 2022 at 14:55

Hint :

Using :

$$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$$

And $$(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca$$

We got :

$$8-6(ab+bc+ca)+5abc\leq 2$$

You can conclude using uvw's method .

• I don't think this way works, I tried this but didn't get the end result. Commented Jun 9, 2022 at 10:59
• @SuzuHirose No it works or you are not aware about the uvw's method . Commented Jun 9, 2022 at 11:00
• Yeah, I didn't get there from this, probably my mistake. River Li's method worked out though. Commented Jun 9, 2022 at 11:01