<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] 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!
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