6. For positive real numbers $a, b, c$ it holds that $a^{2}+b^{2}+c^{2}+a b+b c+c a \leqq 1$. Find the maximum possible value of the sum $a+b+c$.
SOLUTION. When solving this problem, it is useful to look for a relationship between the two given expressions. The first one, $a^{2}+b^{2}+c^{2}+a b+b c+c a$, contains the squares of the variables, while the second, $a+b+c$, only contains the first powers. However, by squaring the second expression, we get

$$
(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2(a b+b c+a c).
$$

The expressions $a^{2}+b^{2}+c^{2}$ and $a b+b c+a c$ are also related, specifically,

$$
a b+b c+a c \leqq a^{2}+b^{2}+c^{2},
$$

which can be easily derived from the obvious inequality $(a-b)^{2}+(b-c)^{2}+(c-a)^{2} \geqq 0$.

The obtained inequality (2) allows us, together with the given inequality, to write

$$
2(a b+b c+a c) \leqq a^{2}+b^{2}+c^{2}+a b+b c+a c \leqq 1
$$

so that

$$
a b+b c+a c \leqq \frac{1}{2}
$$

When we cleverly substitute into equation (1), while also using the given inequality, we get

$$
\begin{aligned}
(a+b+c)^{2} & =a^{2}+b^{2}+c^{2}+2(a b+b c+a c)= \\
& =a^{2}+b^{2}+c^{2}+(a b+b c+a c)+(a b+b c+a c) \leqq 1+\frac{1}{2}=\frac{3}{2}.
\end{aligned}
$$

Since the numbers $a, b, c$ are positive, we obtain the estimate

$$
a+b+c \leqq \sqrt{\frac{3}{2}}
$$

It remains to show that this value can also be achieved. The simplest option is to try $a=b=c$ (since then inequality (2) turns into an equality). In this case, from the given inequality, we get $6 a^{2} \leqq 1$ with equality for $a=b=c=1 / \sqrt{6}$, for which values indeed $a+b+c=\sqrt{3 / 2}$.

GUIDING AND ADDITIONAL PROBLEMS:

N1. For positive numbers $a, b, c$, it holds that $a^{2}+b^{2}+a b \leqq 2$. Find the largest possible value of the product $a b$. [Use the estimate $2 a b \leqq a^{2}+b^{2}$, the largest value $a b=2 / 3$ is achieved for $a=b=\sqrt{2 / 3}$.]

N2. Prove the inequality $a^{2}+b^{2}+c^{2} \geqq a b+b c+c a$ and determine when equality holds. [Multiply by two and rearrange to the form $(a-b)^{2}+(b-c)^{2}+(c-a)^{2} \geqq 0$, equality holds precisely when $a=b=c$.]

N3. Real numbers $a, b, c$ have a sum of 3. Prove that $3 \geqq a b+b c+c a$. When does equality hold? [It follows from the equality $9=(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2(a b+b c+c a)$ and the previous problem. Equality holds only in the case $a=b=c=1$.]

D1. Let $a, b, c$ be positive real numbers whose sum is 3, and each of them is at most 2. Prove that the inequality $a^{2}+b^{2}+c^{2}+3 a b c<9$ holds. 68-C-I-3

D2. For non-negative real numbers $a, b$, it holds that $a^{2}+b^{2}=1$. Determine the smallest and largest possible value of the expression $V=\left(a^{4}+b^{4}+a b+1\right) /(a+b)$. 68-B-I-4