Show that $(b_1+b_2)^2\leq (a_1+a_2)(c_1+c_2)$ How to show that if $a_k,b_k,c_k\in \mathbb{R}^+$ for which $a_kc_k-b_k^2\geq 0$ for $k=1,2$, then
$$
(b_1+b_2)^2\leq (a_1+a_2)(c_1+c_2)?
$$
I tried to check some well-known identities such as Young's inequality, but I failed to make it work.
 A: Conditions imply that quadratics $a_kx^2+2b_kx+c_k$ are nonnegative for all real $x$. Thus, their sum $(a_1+a_2)x^2+2(b_1+b_2)x+(c_1+c_2)$ is nonnegative for all real $x$ as well. Hence, the discriminant of this quadratic is nonpositive and we get the desired inequality.
A: HINT
You can start with the RHS as follows
\begin{align*}
(a_{1} + a_{2})(c_{1} + c_{2}) & = a_{1}c_{1} + a_{1}c_{2} + a_{2}c_{1} + a_{2}c_{2}\\\\
& \geq b^{2}_{1} + a_{1}c_{2} + a_{2}c_{2} + b^{2}_{2}
\end{align*}
Now you can apply the AM-GM inequality in order to obtain
\begin{align*}
a_{1}c_{2} + a_{2}c_{1} & \geq 2\sqrt{a_{1}c_{2}a_{2}c_{1}}\\\\
& = 2\sqrt{a_{1}c_{1}a_{2}c_{2}}\\\\
& \geq 2\sqrt{b^{2}_{1}b^{2}_{2}}\\\\
& = 2b_{1}b_{2}
\end{align*}
Can you take it from here?
A: By the inequality of Cauchy-Schwarz :
$$\begin{array}{lcl}
(a_1 + a_2)(c_1 + c_2) & = & \left(\sqrt{a_1}^2 + \sqrt{a_2}^2\right) \left(\sqrt{c_1}^2 + \sqrt{c_2}^2\right) \\[3mm]
& \geq & \left(\sqrt{a_1 c_1} + \sqrt{a_2 c_2}\right)^2 \\[3mm]
& \geq & \left(\sqrt{b_1^2} + \sqrt{b_2^2}\right)^2 \\[3mm]
& = & \left(b_1 + b_2\right)^2 
\end{array}$$
