It's of course equivalent to \[\lambda \ge \frac{\sqrt{\frac{a^2+b^2}{2}}-\frac{a+b}{2}}{\sqrt{\frac{a^2+b^2}{2}}-\sqrt{ab}}=\frac{1}{2} \cdot \frac{\sqrt{\frac{a^2+b^2}{2}}+\sqrt{ab}}{\sqrt{\frac{a^2+b^2}{2}}+\frac{a+b}{2}}\]So $a=b=1$ implies $\lambda \ge \frac{1}{2}$ and the inequality for $\lambda=\frac{1}{2}$ holds of course by AM-GM.

Tintarn wrote: It's of course equivalent to \[\lambda \ge \frac{\sqrt{\frac{a^2+b^2}{2}}-\frac{a+b}{2}}{\sqrt{\frac{a^2+b^2}{2}}-\sqrt{ab}}=\frac{1}{2} \cdot \frac{\sqrt{\frac{a^2+b^2}{2}}+\sqrt{ab}}{\sqrt{\frac{a^2+b^2}{2}}+\frac{a+b}{2}}\]So $a=b=1$ implies $\lambda \ge \frac{1}{2}$ and the inequality for $\lambda=\frac{1}{2}$ holds of course by AM-GM. Good. Because $ \frac{\sqrt{\frac{a^2+b^2}{2}}+\sqrt{ab}}{\sqrt{\frac{a^2+b^2}{2}}+\frac{a+b}{2}}\leq1$ , so $\lambda_{min} =\frac{1}{2}.$ If $a, b$ be positive real numbers ,then $$a+b\geq\sqrt{ab}+\sqrt{\frac{a^2+b^2}{2}}$$

Find the smallest possible value of the nonnegative number $\lambda$ such that the inequality $$\frac{a+b+c}{3}\geq\lambda \sqrt[3]{abc}+(1-\lambda )\sqrt[3]{\frac{a^3+b^3+c^3}{3}}$$for all positive real numbers $a, b,c$ holds.

$\lambda_{min}=0.68713...$

arqady wrote: $\lambda_{min}=0.6872...$ Thank arqady.

Tintarn wrote: It's of course equivalent to \[\lambda \ge \frac{\sqrt{\frac{a^2+b^2}{2}}-\frac{a+b}{2}}{\sqrt{\frac{a^2+b^2}{2}}-\sqrt{ab}}=\frac{1}{2} \cdot \frac{\sqrt{\frac{a^2+b^2}{2}}+\sqrt{ab}}{\sqrt{\frac{a^2+b^2}{2}}+\frac{a+b}{2}}\]So $a=b=1$ implies $\lambda \ge \frac{1}{2}$ and the inequality for $\lambda=\frac{1}{2}$ holds of course by AM-GM. However you can't take a=b=1 becuase then $\sqrt{\frac{a^2+b^2}{2}}-\sqrt{ab}$ = 0, so then the denominator of the first fraction is 0. You can't multiply both numerator and denominator by 0. Subbing a=b=1 into the original inequality yields 1$\ge$1, which is clearly true for all $\lambda$, and thus this does not show that $\lambda$ is greater than or equal to 1/2. I think the inequality should use limits...

noobatron3000 wrote: However you can't take a=b=1 becuase then $\sqrt{\frac{a^2+b^2}{2}}-\sqrt{ab}$ = 0, so then the denominator of the first fraction is 0. You can't multiply both numerator and denominator by 0. Subbing a=b=1 into the original inequality yields 1$\ge$1, which is clearly true for all $\lambda$, and thus this does not show that $\lambda$ is greater than or equal to 1/2. I think the inequality should use limits... Technically, this is of course correct. A priori, the inequality \[\lambda \ge \frac{1}{2} \cdot \frac{\sqrt{\frac{a^2+b^2}{2}}+\sqrt{ab}}{\sqrt{\frac{a^2+b^2}{2}}+\frac{a+b}{2}}\]is only true when $a \ne b$. But then we see that it is a continuous function also at $a=b$ and hence the inequality must also hold there, so we are allowed to do that (of course you could also work with the limit). After all, the whole point of the problem is the fact that the equality case somehow is $a=b$ but plugging this into the original inequality does not yield anything interesting, as you correctly pointed out.

sqing wrote: Find the smallest possible value of the nonnegative number $\lambda$ such that the inequality $$\frac{a+b}{2}\geq\lambda \sqrt{ab}+(1-\lambda )\sqrt{\frac{a^2+b^2}{2}}$$for all positive real numbers $a, b$ holds. See here too : https://www.cut-the-knot.org/Optimization/HongKong.shtml

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Find the smallest possible value of the nonnegative number $\lambda$ such that the inequality$$\frac{a_1+a_2+\cdots+a_n}{n}\geq\lambda \sqrt[n]{a_1a_2\cdots a_n}+(1-\lambda )\sqrt[n]{\frac{a_1^n+a_2^n+\cdots+a_n^n}{n}}$$for all positive real numbers $a_1,a_2,\cdots,a_n (n\ge 2) $ holds.

Maybe usefull By weighted $AM-GM$ $$\sqrt[n]{\lambda \cdot a_1a_2\cdots a_n+(1-\lambda )\cdot \frac{a_1^n+a_2^n+\cdots+a_n^n}{n} } \ge \lambda \sqrt[n]{a_1a_2\cdots a_n}+(1-\lambda )\sqrt[n]{\frac{a_1^n+a_2^n+\cdots+a_n^n}{n}}$$

arqady wrote: $\lambda_{min}=0.68713...$ How you find this?

Adilet160205 wrote: Maybe usefull By weighted $AM-GM$ $$\sqrt[n]{\lambda \cdot a_1a_2\cdots a_n+(1-\lambda )\cdot \frac{a_1^n+a_2^n+\cdots+a_n^n}{n} } \ge \lambda \sqrt[n]{a_1a_2\cdots a_n}+(1-\lambda )\sqrt[n]{\frac{a_1^n+a_2^n+\cdots+a_n^n}{n}}$$ This is easy, because we consider the function $f(t)=\sqrt[n]{t}$, this is concave, so we can write Jensen: $f(\lambda x+(1-\lambda )y)\ge \lambda f(x)+(1-\lambda )f(y)$

Tintarn wrote: It's of course equivalent to \[\lambda \ge \frac{\sqrt{\frac{a^2+b^2}{2}}-\frac{a+b}{2}}{\sqrt{\frac{a^2+b^2}{2}}-\sqrt{ab}}=\frac{1}{2} \cdot \frac{\sqrt{\frac{a^2+b^2}{2}}+\sqrt{ab}}{\sqrt{\frac{a^2+b^2}{2}}+\frac{a+b}{2}}\]So $a=b=1$ implies $\lambda \ge \frac{1}{2}$ and the inequality for $\lambda=\frac{1}{2}$ holds of course by AM-GM. If a=b, you're dividing by 0 when you go from one expression to another

IMD2 wrote: Tintarn wrote: It's of course equivalent to \[\lambda \ge \frac{\sqrt{\frac{a^2+b^2}{2}}-\frac{a+b}{2}}{\sqrt{\frac{a^2+b^2}{2}}-\sqrt{ab}}=\frac{1}{2} \cdot \frac{\sqrt{\frac{a^2+b^2}{2}}+\sqrt{ab}}{\sqrt{\frac{a^2+b^2}{2}}+\frac{a+b}{2}}\]So $a=b=1$ implies $\lambda \ge \frac{1}{2}$ and the inequality for $\lambda=\frac{1}{2}$ holds of course by AM-GM. If a=b, you're dividing by 0 when you go from one expression to another He used the limit...