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?
 A: Knowing the answer, it is not hard to construct the right CS inequality to solve your problem:
$$52\cdot (13^2+12^2+18^2) \ge (13a+12b+18c)^2 = (7a+6\cdot 26)^2 \implies a \le \frac{26}7$$
It is easy to check equality is attained when $a = \frac{26}7, b = \frac{24}7, c = \frac{36}7$. 

If you don't know the answer to start with, your method can be modified a bit to give the result.  We start with
$$52 \cdot \left(x^2+y^2+z^2 \right) \ge (x a + y b + z c)^2 \tag{1}$$
As the RHS needs to be of form $\left(\alpha a+\beta(a+2b+3c)\right)^2$ to isolate $a$, so we let $y = 2, z = 3$.  Thus we have  
$$52 \cdot \left(x^2+4+9 \right) \ge \left(x a + 2 b + 3 c \right)^2 =\left((x-1) a + 26 \right)^2$$
Equality conditions give
$$\frac{a}x=\frac{b}2=\frac{c}3 \implies b = \frac{2a}x, c = \frac{3a}x$$
and the initial conditions give
$$a=26\frac{x}{x+13}, \quad a^2 = 52\frac{x^2}{x^2+13} \implies x = \frac{13}6, a=\frac{26}7$$
A: The issue is that you are asked to find the least upper bound.
All that you have done is to find an upper bound. You still have yet to show that it is indeed the least upper bound.
In your CS inequality, in order for equality to hold, we must have
$$ \frac{a}{2} = \frac{b}{2} = \frac{ c}{3} , a = 2 \sqrt{221} - 26 $$
However, you can check that the initial conditions would not hold.
A: 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.
A: Our conditions give
$$a^2+b^2+\left(\frac{26-a-2b}{3}\right)^2=52$$ or
$$13b^2-2(52-2a)b+10a^2-52a+208=0.$$
Hence, $$(52-2a)^2-13(10a^2-52a+208)\geq0$$ or
$$a(26-7a)\geq0$$ or
$$0\leq a\leq\frac{26}{7}.$$
For $a=\frac{26}{7}$ we obtain $b=\frac{52-2\cdot\frac{26}{7}}{13}=\frac{20}{7}$ and $c=\frac{26-\frac{26}{7}-2\cdot\frac{20}{7}}{3}>0,$
which says that the answer is $\frac{26}{7}$.
Done!
