Let $a, b, c$ be positive real numbers such that $a + 2b + 3c = 26$ and $a^2 + b^2 + c^2 = 52$. Find the largest possible value of $a$.
I used the Cauchy Schwarz inequality $(ax+by+cz)^2 \leq (a^2+b^2+c^2)(x^2+y^2+z^2)$ as follows:
$a + 2b + 3c = 26$ is given; adding $a$ to both sides gives $2a + 2b + 3c = 26+a$. Then, we have $x = 2$, $y=2$, $z=3$. Putting this all into the inequality, we get:
$(ax+by+cz)^2 \leq (a^2+b^2+c^2)(x^2+y^2+z^2)$
$(2a+2b+3c)^2 \leq (a^2+b^2+c^2)(2^2+2^2+3^2)$
$(26+a)^2 \leq (52)(17)$
$(26+a)^2 \leq 884$
$26 + a \leq \sqrt{884}$
$a \leq 2\sqrt{221} - 26 \approx 3.73$
However, the listed answer gives $a \leq \frac{26}{7} \approx 3.71$. What am I doing wrong?

You can solve $a + 2b + 3c = 26$ for $b$ and plug into $a^2 + b^2 + c^2 = 52$ which gives:
$$13c^2+6ac-156c+5a^2-52a+676=208 \tag{A}$$
Which is a conic section (ellipse) and the maximum can be found by taking a derivative wrt $c$:
$$26c + 6a + 6c\frac{da}{dc} - 156 + 10a\frac{da}{dc} - 52\frac{da}{dc} = 0$$
and setting $\frac{da}{dc} = 0$ :
$$26c + 6a - 156 = 0 \tag{B}$$
Solving (B) for $c$ and plugging into (A) gives:
$$a = \frac{27}6$$
as the max.
