Basic Inequality - 2

touched the 340 mark!!!. anyways, the solution:

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by Cauchy- Schwarz inequity: $(3^2+4^2)(b^2+c^2)\geq (3b+4c)^2$ from the equatins we have $\begin{cases} a^2+b^2+c^2=25\Longrightarrow b^2+c^2=25-a^2\\ a+3b+4c=25\Longrightarrow 3b+4c=25-a\end{cases}$ . substitute these values to find $25(25-a^2)\geq (25-a)^2$ $-25a^2+625\geq a^2-50a+625$ $0\geq 26a^2-50a=26a(a-\dfrac{50}{26})$ this is just a quadratic inequity which gives us: $0\leq a\leq \dfrac{50}{26}\approx \boxed{1.923}$

It is obvious that $3b+4c = 25 - a$ and $b^{2} + c^{2} = 25 - a^{2}$ . By using Cauchy-Schwarz Inequality we have

$(b^{2} + c^{2})(9 + 16) \geq (3b + 4c)^{2} <=> b^{2} + c^{2} \geq \frac{(3b +4c)^{2}}{25}$ .

Therefore,

$25 - a^{2} = b^{2} + c^{2} \geq \frac{(25-a)^{2}}{25} <=> 625 - 25a^{2} \geq (25-a)^{2} <=> 26a^{2} - 50a \leq 0 <=> a \in [0, \frac{25}{13}]$ .