This problem is proposed by Alfreed and me!

WE ARE MAKING IT OUT OF THE CALCULUS CLASS WITH THIS ONE, FR, FR!!1!1!. Claim 1: $\alpha \ge \frac{1}{2}$ Proof: Note that by setting $x=1+a$ and $y=1-a$ we can get $\alpha \ge \frac{\sqrt{a^2+1}-1}{\sqrt{a^2+1}-\sqrt{1-a^2}}=f(a)$, so now we are gonna compute $\lim_{a \to 0} f(a)$ which i claim it is $\frac{1}{2}$, indeed, by L'hopital: $$\lim_{a \to 0} \frac{\sqrt{a^2+1}-1}{\sqrt{a^2+1}-\sqrt{1-a^2}}=\lim_{a \to 0} \dfrac{\frac{a}{\sqrt{a^2+1}}}{\frac{a}{\sqrt{a^2+1}}+\frac{a}{\sqrt{1-a^2}}}=\lim_{a \to 0} \dfrac{\frac{1}{\sqrt{a^2+1}}}{\frac{1}{\sqrt{a^2+1}}+\frac{1}{\sqrt{1-a^2}}}=\frac{1}{1+1}=\frac{1}{2}$$From here we can clearly say $\alpha \ge \frac{1}{2}$ as desired. Finishing: All we need to do is to prove $\alpha=\frac{1}{2}$ actually works, first note that we can fix the sum $x+y=k$, and since the inequality is homogeneous, we can scale pairs $(x,y) \to (zx, zy)$ hence we can WLOG $x+y=1$, let $t=x-x^2$, notice that $f(x)=x-x^2$ satisfies $f'(x)=1-2x$ so its peak point in the range $(0,1)$ is $x=\frac{1}{2}$ meaning that $\frac{1}{4} \ge t \ge 0$. The ineq is now: $$1 \ge \sqrt{t}+\sqrt{\frac{1}{2}-t}=f(t) \; \text{if and only if} \; f'(t)=\frac{1}{2\sqrt{t}}-\dfrac{1}{2\sqrt{\frac{1}{2}-t}} \ge 0 \; \forall \frac{1}{4} \ge t \ge 0$$The last ineq holds clearly holds, the reason why the if and only if, is because $f(t)$ is crescent in the range of $t$ with maximun when $t=\frac{1}{4}$, so we get $1=f \left(\frac{1}{4} \right) \ge f(t)$ as desired, thus we are done .

Here's a one-line proof of the inequality for $\alpha= \frac{1}{2}.$ Substitute $A=\sqrt{xy}$ and $B=\sqrt{\frac{x^2+y^2}{2}}$ and the inequality becomes $\sqrt{\frac{A^2+B^2}{2}} \ge \frac{A+B}{2}.$

Filipjack wrote: Here's a one-line proof of the inequality for $\alpha= \frac{1}{2}.$ Substitute $A=\sqrt{xy}$ and $B=\sqrt{\frac{x^2+y^2}{2}}$ and the inequality becomes $\sqrt{\frac{A^2+B^2}{2}} \ge \frac{A+B}{2}.$ cleeeeeean

Here is a straightforward epsilon- and analysis free solution to the problem. The main virtue of this approach, however, is in my opinion that it does not rely on guessing the correct answer or equality case. W.l.o.g. $\alpha \le 1$. By homogeneity we may w.l.o.g. put $y=1$ and $x=t^2$ so that the question becomes when \[t^2-2\alpha t+1 \ge (1-\alpha)\sqrt{2+2t^4}.\]Squaring and rearranging we get the equivalent \[(2\alpha^2-4\alpha+1)t^4+4\alpha t^3-(4\alpha^2+2)t^2+4\alpha t+(2\alpha^2-4\alpha+1) \ge 0.\]From the original problem we know that $x=y$ is an equality case for all $\alpha$ and hence here we know that $t=1$ must be a double root for all $\alpha$. Hence we can factor the LHS as \[(t-1)^2((2\alpha^2-4\alpha+1)t^2+(4\alpha^2-4\alpha+2)t+(2\alpha^2-4\alpha+1) \ge 0.\]Hence we must have \[(2\alpha^2-4\alpha+1)t^2+(4\alpha^2-4\alpha+2)t+(2\alpha^2-4\alpha+1 \le 0\]for all $t \ne 1$ and hence for all $t$. Now we could just plug in $t=1$ to see that $\alpha \ge\frac{1}{2}$ is necessary, and conversely $\alpha=\frac{1}{2}$ to get the true $-\frac{1}{2}(t-1)^2 \le 0$. However, we don't even need to make a guess, as we have a quadratic function of $t$, so we can just compute its discriminant to be $16\alpha(2\alpha-1)(\alpha-1)$ from where it is immediate that the correct condition is $\alpha \ge \frac{1}{2}$.

It is worth mentioning the same problem was proposed as the first problem of the second day(5th problem) of the 9th grade Belarus National Olympiad 2014.

Actually, it is quite funny how the last algebra problem of Baltic Way is the first for 9-graders in Belarus : ). Hope Estonia will catch up with Belarus!

Well, it deserves to be mentioned that the problems in Baltic Way are by no means intended to be ordered by difficulty. In fact, the results show that Problems 2 and 4 were harder than Problem 5...

a_507_bc wrote: Find the smallest positive real $\alpha$, such that $$\frac{x+y} {2}\geq \alpha\sqrt{xy}+(1 - \alpha)\sqrt{\frac{x^2+y^2}{2}}$$for all positive reals $x, y$. Hong Kong TST, 2017. A few solutions (including mine), below https://www.cut-the-knot.org/Optimization/HongKong.shtml?fbclid=IwAR23_QCyvmFCkaHZPZ3ruhjOj9YEHZ2MraVa_4Y-BS7UB2q4YqXgQvlxAD8

Bump for another proofs, of course

mihaig wrote: Bump for another proofs, of course Well, we have already seen at least three very different proofs. Unless you are asking for a very specific new proof, I don't see a point in bumping this...

$\frac{x+y} {2}\geq \alpha\sqrt{xy}+(1 - \alpha)\sqrt{\frac{x^2+y^2}{2}}$ $\alpha (\sqrt{2(x^2+y^2)}-2\sqrt{xy})\ge \sqrt{2(x^2+y^2)}- (x+y)$ $\alpha \frac{2(x^2+y^2)-4xy}{\sqrt{2(x^2+y^2)}+2\sqrt{xy}}\ge \frac{2(x^2+y^2)-(x+y)^2}{\sqrt{2(x^2+y^2)}+ x+y}$ $\frac{(x-y)^2}{\sqrt{2(x^2+y^2)}+2\sqrt{xy}} \big(2\alpha - \frac{\sqrt{2(x^2+y^2)}+2\sqrt{xy}}{\sqrt{2(x^2+y^2)}+ x+y} \big)\ge 0$ $\alpha_{min}=1/2 \cdot \max\big( \frac{\sqrt{2(x^2+y^2)}+2\sqrt{xy}}{\sqrt{2(x^2+y^2)}+ x+y}\big)=1/2$

On the one hand, the inequality can be rewritten as $$\frac{(x-y)^2}{4}\left(\frac{\left(\sqrt{x}-\sqrt{y}\right)^2}{\sqrt{2\left(x^2+y^2\right)}+x+y}+(2\alpha -1)\right)\ge 0,$$and therefore if $\alpha\ge\frac{1}{2}$ then the inequality holds for all $x,y>0$. On the other hand, suppose there exists an $\alpha=\frac{1}{2}t$, where $0<t<1$, such that the main inequality holds for all $x,y>0$. In particular, it must hold for $x=t^3$ and $y=t$. But then $$\frac{x+y}{2}-\alpha\sqrt{xy}-(1-\alpha)\sqrt{\frac{x^2+y^2}{2}}=\frac{t}{2\sqrt{2}}\cdot\frac{(t-1)^3\left(2+2t+t^2-t^3\right)}{\sqrt{2}+(2-t)\sqrt{t^4+1}}<0$$for all $t$, a contradiction.

Tintarn wrote: mihaig wrote: Bump for another proofs, of course Well, we have already seen at least three very different proofs. Unless you are asking for a very specific new proof, I don't see a point in bumping this... We also have already seen the problem. And also your proof. 7 years ago

mihaig wrote: We also have already seen the problem. And also your proof. 7 years ago So? If you even think my solution was superfluous, then I understand even less why you are bumping to ask for more proofs...

Tintarn wrote: mihaig wrote: We also have already seen the problem. And also your proof. 7 years ago So? If you even think my solution was superfluous, then I understand even less why you are bumping to ask for more proofs... You are the one who begun. So let's stop here