minimum value of $l$

Question

Minimum positive real number $l$ for which

$7\sqrt{a}+17\sqrt{b}+l\sqrt{c}\geq 2019.$ given that $a+b+c=1$ and $a,b,c>0$

what i try

cauchy Inequality

$$(7^2+17^2+l^2)(a+b+c)\geq \bigg(7\sqrt{a}+17\sqrt{b}+l\sqrt{c}\bigg)^2$$

How do i solve it. Help me please

Answer 1

Now, solve the following inequality. $$7^2+17^2+l^2\geq2019^2.$$

Answer 2

From $(7^2+17^2+l^2)(a+b+c)\geq \bigg(7\sqrt{a}+17\sqrt{b}+l\sqrt{c}\bigg)^2$ you get

$(7^2+17^2+l^2)(a+b+c)\geq 2019^2$. Since $a+b+c=1$ we get

$7^2+17^2+l^2\geq 2019^2$

Can you proceed ?