<s>[INST] 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?
 [/INST] 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.
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