$M_1^2+M_2^2+M^2_3=M$ ,where $M_1,M_2,M_3$ are positive numbers and sum of any two is greater than the third, show that $2M ≤ (M_1 +M_2 +M_3)^2 ≤ 3M.$

Question

$M_1^2+M_2^2+M^2_3=M$ ,where $M_1,M_2,M_3$ are positive numbers and sum of any two is greater than the third, show that $2M ≤ (M_1 + M_2 + M_3)^2 ≤ 3M.$

My solution goes like this:

We know that, $M_1^2+M_2^2+M^2_3\geq \Sigma M_1M_2$. Thus, $(M_1+M_2+M^2)^2=M_1^2+M_2^2+M^2_3+2\Sigma M_1M_2\leq 3M$

But I dont think, $2M ≤ (M_1 + M_2 + M_3)^2$ is true. For then, $M\leq 2\Sigma M_1M_2$, a contradiction. Is the first inequality, valid? Also, is my solution correct?

Answer 1

The one on the right is CS inequality, for the one on the left. Use: $M_1+M_2 > M_3 \implies M_1M_3+M_2M_3 > M_3^2$. Do the same for two more inequalities of the same type and add them up. Then add both sides to $M_1^2+M_2^2+M_3^2$ to complete the square.