$f(x)=\sqrt{a\sin^2(x) +b\cos^2(x)+c} +\sqrt{a\cos^2(x)+b\sin^2(x)+c}$, Prove that $\sqrt{a+ c} + \sqrt{b + c} \le f(x) \le 2\sqrt{\frac{a+ b}{2}+c}$

Question

$f(x) = \sqrt{a\sin^2(x) + b\cos^2(x) + c} + \sqrt{a\cos^2(x) + b\sin^2(x) + c}$

$a, b, c \in \mathbb{R}$

Answer 1

The right inequality.

Since $g(x)=\sqrt{x}$ is a concave function, by Jensen we obtain: $$f(x)\leq2\sqrt{\frac{a\sin^2x+b\cos^2x+c+a\cos^2x+b\sin^2x+c}{2}}=2\sqrt{\frac{a+b}{2}+c}.$$ The left inequality.

We need to prove that: $$a+b+2c+2\sqrt{(a\sin^2x+b\cos^2x+c)(a\cos^2x+b\sin^2x+c)}\geq$$ $$\geq a+b+2c+2\sqrt{(a+c)(b+c)}$$ or $$(a\sin^2x+b\cos^2x+c)(a\cos^2x+b\sin^2x+c)\geq(a+c)(b+c)$$ or $$(a^2+b^2)\sin^2x\cos^2x+ab(\sin^4x+\cos^4x-1)\geq0$$ or $$(a^2+b^2)\sin^2x\cos^2x+ab(\sin^4x+\cos^4x-(\sin^2x+\cos^2x)^2)\geq0$$ or $$(a-b)^2\sin^2x\cos^2x\geq0,$$ which is obvious.

Answer 2

Another way.

The right inequality.

By C-S $$f(x)\leq\sqrt{(1+1)(a\sin^2x+b\cos^2x+c+a\cos^2x+b\sin^2x+c)}=$$ $$=\sqrt{2(a+b+2c)}=2\sqrt{\frac{a+b}{2}+c}.$$ The left inequality.

Since $f(x)=f(-x)$ and $f\left(\frac{\pi}{2}-x\right)=f(x),$ we can assume that $x\in\left[0,\frac{\pi}{4}\right],$

which gives $\cos2x\geq0.$

Also, we can assume that $a\geq b.$

Now, after squaring of the both sides(see my previous solution) we need to prove that $$(a\sin^2x+b\cos^2x+c)(a\cos^2x+b\sin^2x+c)\geq(a+c)(b+c)$$ or $$\ln(a\cos^2x+b\sin^2x+c)+\ln(a\sin^2x+b\cos^2x+c)\geq\ln(a+c)+\ln(b+c),$$ which is true by Karamata because $$(a+c,b+c)\succ(a\cos^2x+b\sin^2x+c,a\sin^2x+b\cos^2x+c)$$ and $\ln$ is a concave function.

Also, the left inequality i's just Karamata for the concave function $h(x)=\sqrt{x}.$

Answer 3

Similar to what this answer does, use the rearrangements of the double-angle formulae for $\cos(2x)$ to get

$$\sin^2(x)=\frac{1-\cos(2x)}{2}, \; \; \cos^2(x)=\frac{1+\cos(2x)}{2} \tag{1}\label{eq1A}$$

which then leads to

$$\begin{equation}\begin{aligned} f(x) & = \sqrt{a\sin^2(x) + b\cos^2(x) + c} + \sqrt{a\cos^2(x) + b\sin^2(x) + c} \\ & = \sqrt{a\left(\frac{1-\cos(2x)}{2}\right) + b\left(\frac{1+\cos(2x)}{2}\right) + c} \;\; + \\ & \;\;\;\; \sqrt{a\left(\frac{1+\cos(2x)}{2}\right) + b\left(\frac{1-\cos(2x)}{2}\right) + c} \\ & = \sqrt{\left(\frac{a+b}{2}+c\right) + \left(\frac{b-a}{2}\right)\cos(2x)} \;\; + \\ & \;\;\;\; \sqrt{\left(\frac{a+b}{2}+c\right) - \left(\frac{b-a}{2}\right)\cos(2x)} \\ & = \sqrt{A + B\cos(2x)} + \sqrt{A - B\cos(2x)} \end{aligned}\end{equation}\tag{2}\label{eq2A}$$

where

$$A = \frac{a+b}{2}+c, \; \; B = \frac{b-a}{2} \tag{3}\label{eq3A}$$

Similar to what you did, taking the derivative of $f(x)$ and setting it to $0$ gives

$$\begin{equation}\begin{aligned} 0 & = \frac{-B\sin(2x)}{\sqrt{A+B\cos(2x)}}+\frac{B\sin(2x)}{\sqrt{A-B\cos(2x)}} \\ & = B\sin(2x)\left(\frac{-1}{\sqrt{A+B\cos(2x)}}+\frac{1}{\sqrt{A-B\cos(2x)}}\right) \end{aligned}\end{equation}\tag{4}\label{eq4A}$$

There are now several possibilities. First, if $B=0$, then from \eqref{eq3A} we have $a=b$, so $f(x) = \sqrt{a+c} + \sqrt{b+c}$. Thus, it's equal to the your LHS minimum value.

With $B \neq 0$, the next possibility is that $\sin(2x)=0$, so $\cos(2x)=\pm 1$. This gives a minimum (which can be confirmed by checking the second derivative) of $f(x) = \sqrt{a+c} + \sqrt{b+c}$, i.e., your LHS inequality.

The final possibility is that

$$\begin{equation}\begin{aligned} 0 & = \frac{-1}{\sqrt{A+B\cos(2x)}}+\frac{1}{\sqrt{A-B\cos(2x)}} \\ \frac{1}{\sqrt{A+B\cos(2x)}} & = \frac{1}{\sqrt{A-B\cos(2x)}} \\ A-B\cos(2x) & = A+B\cos(2x) \\ \cos(2x) & = 0 \end{aligned}\end{equation}\tag{5}\label{eq5A}$$

This gives a maximum, with using \eqref{eq2A} and \eqref{eq3A} giving the below, i.e., the RHS of your inequality

$$f(x) = 2\sqrt{\frac{a+b}{2}+c} \tag{6}\label{eq6A}$$

Answer 4

My other answer's $(2)$ and $(3)$ determine

$$f(x) = \sqrt{A + B\cos(2x)} + \sqrt{A - B\cos(2x)} \tag{1}\label{eq1B}$$

$$A = \frac{a+b}{2}+c, \; \; B = \frac{b-a}{2} \tag{2}\label{eq2B}$$

To prove the LHS inequality, use \eqref{eq2B} to get

$$\begin{equation}\begin{aligned} (A+B\cos(2x))(A-B\cos(2x)) & = A^2 - B^2\cos^2(2x) \\ & \ge A^2 - B^2 \\ & = \left(\frac{a+b}{2}+c\right)^2 - \left(\frac{b-a}{2}\right)^2 \\ & = ab + ac + bc + c^2 \\ & = (a+c)(b+c) \end{aligned}\end{equation}\tag{3}\label{eq3B}$$

With your original $f(x)$ expression and \eqref{eq1B}, this leads to

$$\begin{equation}\begin{aligned} a+b+2c+2\sqrt{A+B\cos(2x)}\sqrt{A-B\cos(2x)} & \ge a+b+2c+2\sqrt{a+c}\sqrt{b+c} \\ \left(\sqrt{A+B\cos(2x)} + \sqrt{A-B\cos(2x)}\right)^2 & \ge \left(\sqrt{a+c} + \sqrt{b+c}\right)^2 \\ f(x) & \ge \sqrt{a+c} + \sqrt{b+c} \end{aligned}\end{equation}\tag{4}\label{eq4B}$$

For the RHS inequality, from \eqref{eq3B} we also have $(A+B\cos(2x))(A-B\cos(2x)) \le A^2$, which similar to \eqref{eq4B} and using \eqref{eq2B}, then gives

$$\begin{equation}\begin{aligned} & a+b+2c+2\sqrt{A+B\cos(2x)}\sqrt{A-B\cos(2x)} \le a+b+2c+2\left(\sqrt{\frac{a+b}{2}+c}\right)^2 \\ & \left(\sqrt{A+B\cos(2x)} + \sqrt{A-B\cos(2x)}\right)^2 \le \left(\left(\sqrt{\frac{a+b}{2}+c}\right) + \left(\sqrt{\frac{a+b}{2}+c}\right)\right)^2 \\ & \sqrt{A+B\cos(2x)} + \sqrt{A-B\cos(2x)} \le \sqrt{\frac{a+b}{2}+c} + \sqrt{\frac{a+b}{2}+c} \\ & \qquad\qquad\qquad\qquad\qquad\qquad\qquad f(x) \le 2\sqrt{\frac{a+b}{2}+c} \end{aligned}\end{equation}\tag{5}\label{eq5B}$$

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Alternatively, to get the RHS inequality, first set $g(x) = \sqrt{a\sin^2(x) + b\cos^2(x) + c}$ and $h(x) = \sqrt{a\cos^2(x) + b\sin^2(x) + c}$. Then the inequality between the arithmetic mean and quadratic mean, as shown in HM-GM-AM-QM inequalities, gives that

$$\begin{equation}\begin{aligned} \frac{g(x) + h(x)}{2} & \le \sqrt{\frac{g^2(x) + h^2(x)}{2}} \\ \frac{f(x)}{2} & \le \sqrt{\frac{a+b+2c}{2}} \\ f(x) & \le 2\sqrt{\frac{a+b}{2}+c} \end{aligned}\end{equation}\tag{6}\label{eq6B}$$