Turning discontinuity into continuity (2017 CHKMO)

Since the inequality is homogeneous of degree $2$ , we can assume without loss of generality that $a^2 + b^2 = 2$ , and therefore we can write $a = \sqrt{2}\cos\theta$ and $b = \sqrt{2}\sin\theta$ for $0 \le \theta \le \tfrac12\pi$ . Thus we require $\begin{aligned} \tfrac{1}{\sqrt{2}}(\cos\theta + \sin\theta) & \ge \; \lambda\sqrt{2\sin\theta\cos\theta} + 1 - \lambda \\ \lambda\big(1 - \sqrt{\sin2\theta}\big) & \ge \; 1 - \cos(\theta-\tfrac14\pi) \end{aligned}$ for $0 \le \theta \le \tfrac12\pi$ . Putting $\varphi = \theta - \tfrac14\pi$ , we need $\begin{aligned} \lambda\big(1 - \sqrt{\cos2\varphi}\big) & \ge \; 1 - \cos\varphi \\ \lambda & \ge \; \frac{1 - \cos\varphi}{1 - \sqrt{\cos2\varphi}} \; =\; \frac{(1 - \cos\varphi)\big(1 + \sqrt{\cos2\varphi}\big)}{1 - \cos2\varphi} \; = \; \frac{1 + \sqrt{\cos2\varphi}}{4\cos^2\frac12\varphi} \end{aligned}$ for $-\tfrac14\pi \le \varphi = \theta - \tfrac14\pi \le \tfrac14\pi$ . Putting $c= \cos\varphi$ we require $\lambda \; \ge \; F(c) \; = \; \frac{1 + \sqrt{2c^2 - 1}}{2(c+1)} \hspace{2cm} \tfrac{1}{\sqrt{2}} \le c \le 1$ Now $F'(c) \; =\; \frac{c}{(c+1)\sqrt{2c^2-1}} - \frac{1 + \sqrt{2c^2-1}}{2(c+1)^2} \; = \; \frac{1}{2(c+1)^2\sqrt{2c^2-1}}\Big[2c(c+1) - \sqrt{2c^2-1}\big(1+\sqrt{2c^2-1}\big)\Big] \; =\; \frac{1 + 2c - \sqrt{2c^2-1}}{2(c+1)^2\sqrt{2c^2-1}}$ is positive for $\tfrac{1}{\sqrt{2}} \le c \le 1$ , and hence the maximum value of $F(c)$ is $F(1)$ . Thus the least possible value of $\lambda$ is $F(1) = \boxed{\tfrac12}$ .

Very cool elementary inequalities problem!

We claim that $\boxed{\lambda = 0.5}$ is the smallest possible value. This is easily verifiable to work, as follows:

$\begin{aligned} \frac{a+b}{2} &\stackrel{?}{\ge} \frac 1 2 \left( \sqrt{ab} + \sqrt{\frac{a^2+b^2}{2}}\right)\\ a+b &\stackrel{?}{\ge} \sqrt{ab} + \sqrt{\frac{a^2+b^2}{2}}\\ \end{aligned}$

But this is true by RMS-AM:

$\begin{aligned} \sqrt{\frac{\sqrt{ab}^2 + \sqrt{\frac{a^2+b^2}{2}}}{2}^2} &\ge \frac{\sqrt{ab} + \sqrt{\frac{a^2+b^2}{2}}}{2}\\ \sqrt{\frac{ab + \frac{a^2+b^2}{2} }{2}} &\ge \frac{\sqrt{ab} + \sqrt{\frac{a^2+b^2}{2}}}{2}\\ \sqrt{\frac{\frac{(a^2+2ab+b^2)}{2} }{2}} &\ge \frac{\sqrt{ab} + \sqrt{\frac{a^2+b^2}{2}}}{2}\\ \sqrt{\left(\frac{a+b}{2}\right)^2} &\ge \frac{\sqrt{ab} + \sqrt{\frac{a^2+b^2}{2}}}{2}\\ a+b &\ge \sqrt{ab} + \sqrt{\frac{a^2+b^2}{2}} \blacksquare \end{aligned}$

Now we must show that any smaller $\lambda$ will not work. Note that the equation is $\text{AM} \ge \lambda (\text{GM}) + (1-\lambda)(\text{RMS})$ . It is known that $\text{AM} \ge \text{GM}$ , and $\text{RMS} \ge \text{AM}$ . So for any smaller $\lambda$ , the pair $a = 1$ and $b = \displaystyle \lim_{x \rightarrow 0}{1+x}$ will not satisfy the inequality for sufficiently small $x$ for each constant $\lambda$ . This can be verified through direct computation.

(x-y)²>=0
=> (x+y)²+(x-y)²>=(x+y)²
=>2(x²+y²)>=(x+y)²
=> $\frac {x²+y²}{2}$ >= $\frac {(x+y)²}{4}$
=> $\sqrt{\frac {x²+y²}{2}}$ >= $\frac {x+y}{2}$
Put x= $\sqrt{\frac {a²+b²}{2}}$ ,y= $\sqrt{ab}$ and we have our result.

Notice that we can write $\lambda \geqslant \frac{\sqrt{\frac{a^2+b^2}{2}}-\frac{a+b}{2}}{\sqrt{\frac{a^2+b^2}{2}}-\sqrt{ab}}=\frac{1}{2} \cdot \frac{\sqrt{\frac{a^2+b^2}{2}}+\sqrt{ab}}{\sqrt{\frac{a^2+b^2}{2}}+\frac{a+b}{2}}$ setting $a=b=1$ , we get $\lambda \geqslant \frac{1}{2}$ . When $\lambda =\frac{1}{2}$ , it's just QM-AM Inequality. $\quad \blacksquare$