Prove that sum of cos/sum of sin ≥ 3 given constraint

[lfdahl]

If $x_1,x_2,...,x_{10} \in [0;\frac{\pi}{2}]$ and $\sum_{i=1}^{10}\sin^2x_i = 1.$

- then prove, that:

\[ \frac{\sum_{i=1}^{10}\cos x_i}{\sum_{j=1}^{10}\sin x_j} \ge 3\]

 

[Albert1]

If $x_1,x_2,...,x_{10} \in [0;\frac{\pi}{2}]$ and $\sum_{i=1}^{10}\sin^2x_i = 1.$

- then prove, that:

\[ \frac{\sum_{i=1}^{10}\cos x_i}{\sum_{j=1}^{10}\sin x_j} \ge 3\]

my solution:

Spoiler

let $p=\sum_{i=1}^{10}\cos x_i=(cos\,{x_1}+----+\cos\,{x_{10}})$

$q=\sum_{i=1}^{10}\sin x_i=(sin\,{x_1}+----+\sin\,{x_{10}})$
so $1\leq q^2\leq 3 ---(1) $ from $---(2)$
for $x_1,x_2,...,x_{10} \in [0;\frac{\pi}{2}]$ and $\sum_{i=1}^{10}\sin^2x_i = 1.---(2)$
we have :$p\geq \sum_{i=1}^{10}\cos^2x_i = 9.$
so $ \dfrac {p}{q}\geq \dfrac{9}{q^2}---(3)$
from (1)(2)(3) we have :$\dfrac {p}{q}\geq 3$

 

[lfdahl]

my solution:

Spoiler

let $p=\sum_{i=1}^{10}\cos x_i=(cos\,{x_1}+----+\cos\,{x_{10}})$

$q=\sum_{i=1}^{10}\sin x_i=(sin\,{x_1}+----+\sin\,{x_{10}})$
so $1\leq q^2\leq 3 ---(1) $ from $---(2)$
for $x_1,x_2,...,x_{10} \in [0;\frac{\pi}{2}]$ and $\sum_{i=1}^{10}\sin^2x_i = 1.---(2)$
we have :$p\geq \sum_{i=1}^{10}\cos^2x_i = 9.$
so $ \dfrac {p}{q}\geq \dfrac{9}{q^2}---(3)$
from (1)(2)(3) we have :$\dfrac {p}{q}\geq 3$

Very well done, Albert! Thankyou for your nice solution!

 

[lfdahl]

An alternative solution:

Spoiler

View attachment 7033