it is given that $x^2 + 2y^2 + 3z^2 = \frac{1}{126}$ (all variables are real numbers)
And i need to prove that it is always the case that: $7x + 10y + 9z \leq 1$
and what I've been able to come up with so far is:
i have defined vector "a" as (x,y,z) and b as (7,10,9). Now i used the two vectors to find the scaler product of them which results in the terms on the left in the statement i am trying to prove. I thought of using the C-S inequality and somehow including the given condition in the inequality to prove the statement. But the problem is the fact that the term 7x + 10y + 9z needs to be in absolute value before the C-S inequality can be used, and I am not sure how to go about this.
$\endgroup$1 Answer
$\begingroup$$$(7x+10y+9z)^2 = (7\cdot x + 5\sqrt{2}\cdot \sqrt{2}y + 3\sqrt{3}\cdot \sqrt{3}z)^2$$$$\le [7^2+(5\sqrt{2})^2+(3\sqrt{3})^2][x^2+2y^2+3z^2]=1$$
by use of Cauchy-Schwarz. I leave for you to complete the proof and show the equality case.
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