Here we say $a+b=x$ and $ab=y$ and our problem will be equialent to $x^2\geq 4y$ which means $a^2+b^2\geq 2ab$ (A.M-G.M)

From $x+y \leq \sqrt{2(x^2+y^2)}$ we get: $ \displaystyle{\sqrt{\dfrac{a^2+ab+b^2}{3}}+\sqrt{ab}\leq \sqrt{2 \Big( \dfrac{a^2+ab+b^2}{3}+ab \Big)}= \sqrt{ \dfrac{2}{3} \Big( a^2+4ab+b^2 \Big)}.} $ It's sufficient to prove that: $ \sqrt{ \dfrac{2}{3} \Big( a^2+4ab+b^2 \Big)} \leq a+b .$ This is equivalent to: $ \dfrac{2}{3} \Big( a^2+4ab+b^2 \Big) \leq a^2+2ab+b^2 \Leftrightarrow 2a^2+8ab+2b^2 \leq 3a^2+6ab+3b^2 \Leftrightarrow 0 \leq a^2-2ab+b^2 \Leftrightarrow 0 \leq (a-b)^2, $ which is obvious.

The following inequality is also true $\boxed{\text{A3}}$If $a,b$ be positive real numbers, show that:$$ \displaystyle{\sqrt{\dfrac{a^2+b^2}{2}}+\sqrt{ab}\leq a+b}$$

(a^2+b^2)/2 >= (a^2+b^2+ab)/3 (easily proven by AM-GM) sqrt(a^2+b^2)/sqrt(2) >=sqrt((a^2+b^2+ab)/3) sqrt(a^2+b^2)/sqrt(2) + sqrt(ab)>=sqrt((a^2+b^2+ab)/3)+sqrt(ab) Our problem becomes proving the inequality a+b >= sqrt(a^2+b^2)/sqrt(2) + sqrt(ab) We can do this by multiplying the whole equality with sqrt(2) and then squaring up: 2a^2+2b^2+4ab >= a^2 + b^2 + 2ab + 2sqrt(2ab)(a^2+b^2) a^2 + b^2 + 2ab >= 2sqrt(2ab)(a^2+b^2) and this is true by AM-GM Of course, on competitions you might want to write from the last one to the top one just to get all the points. Also, if you are not 100% sure about solving this you might want to write this down on a piece of paper.

luofangxiang wrote: The following inequality is also true $\boxed{\text{A3}}$If $a,b$ be positive real numbers, show that:$$ \displaystyle{\sqrt{\dfrac{a^2+b^2}{2}}+\sqrt{ab}\leq a+b}$$ It is C-S.

Brutal way: $$\sqrt{\dfrac{a^2+ab+b^2}{3}}+\sqrt{ab}\le a+b$$$$\iff \dfrac{a^2+4ab+b^2}{3}+2\sqrt{\dfrac{a^3b+a^2b^2+b^3a}{3}}\le a^2+b^2+2ab$$$$\iff \dfrac{2}{3}(a^2+ab+b^2)\ge 2\sqrt{\dfrac{a^3b+a^2b^2+b^3a}{3}}$$$$\iff \dfrac{4}{9}(a^2+ab+b^2)^2\ge \dfrac{4(a^3b+a^2b^2+b^3a)}{3}$$$$\iff a^4+b^4+3a^2b^2+2a^3b+2b^3a\ge 3(a^3b+a^2b^2+b^3a)$$$$\iff a^4+b^4\ge b^3a+a^3b$$Which is true using Muirhead or rearrangment

luofangxiang wrote: The following inequality is also true $\boxed{\text{A3}}$If $a,b$ be positive real numbers, show that:$$ \displaystyle{\sqrt{\dfrac{a^2+b^2}{2}}+\sqrt{ab}\leq a+b}$$ $ 2(x^2+y^2) \geq (x+y)^2 \Leftrightarrow x+y \le \sqrt{2(x^2+y^2)} $ Applying this to the quantity: $ \sqrt { \frac{a^2+b^2}{2} }+\sqrt {ab} \le \sqrt {2(\frac{a^2+2ab+b^2} {2})} = a+b $

If $a,b$ be positive real numbers, then$$ \displaystyle{\sqrt{\dfrac{a^2+ab+b^2}{3}}+\sqrt{ab}\leq \sqrt{\dfrac{a^2+b^2}{2}}+\sqrt{ab}\leq a+b}$$

What was the morality of this act?

My Solution $$ \frac{\sqrt{\frac{a^2+b^2+ab}{3}}+\sqrt{ab}}{2}\leq \sqrt{\frac{(\sqrt{\frac{a^2+b^2+ab}{3}})^2+(\sqrt{ab})^2}{2}}=\sqrt{\frac{1}{6}(a^2+4ab+b^2)}=\sqrt{\frac{1}{6}((a+b)^2+2ab)}\leq \sqrt{\frac{1}{6}((a+b)^2+\frac{(a+b)^2}{2})}=\frac{a+b}{2} $$

We have $\sqrt{\frac{a^2+ab+b^2}{3}}\leq \sqrt{\frac{\frac{3}{2}(a^2+b^2)}{3}}=\sqrt{\frac{a^2+b^2}{2}}$, so we have that the original inequality holds iff \[\sqrt{\frac{a^2+b^2}{2}}\leq a+b-\sqrt{ab} \iff 4\sqrt{ab}(a+b)\leq (a+b)^2+4ab\]which follows from AM-GM.

$$ \displaystyle{\sqrt{\dfrac{a^2+ab+b^2}{3}}+\sqrt{ab}\leq a+b}$$Is equilavent to $LHS\leq \frac{a+b}{2}+\frac{a+b}{2}\leq a+b$

knm2608 wrote: luofangxiang wrote: The following inequality is also true $\boxed{\text{A3}}$If $a,b$ be positive real numbers, show that:$$ \displaystyle{\sqrt{\dfrac{a^2+b^2}{2}}+\sqrt{ab}\leq a+b}$$ It is C-S. I'm afraid that cauchy don't work

the inequality is homogenous so we can consider $a+b=1$ after squaring and subtracting we get that we should prove that: \[(\frac23 - \frac{2 a}{3} + \frac{2 a^2}{3})^2 \ge \frac49 (a - a^2) (3 - 3 a + 3 a^2) \quad (1)\]now take RH to LH we have the identity that $\frac{4 (a^2 - a + 1) (2 a - 1)^2}{9} \ge 0$ and it is equivalent to $(1)$. as desired

Let $s = a+b$ and $p = ab$ Then, the given inequality becomes $\sqrt{\frac{s^2-p}{3}} + \sqrt{p} \le s$, which after rearranging and simplifying, becomes $s^2 + 3p \ge 3s \sqrt{p} \ge 0$, which is just $(s-2\sqrt{p})(s-\sqrt{p}) \ge 0$ But since $s = a+b \ge 2 \sqrt{ab} = 2 \sqrt{p}$, this is true. $\blacksquare$

Looks like I have to do this... Since everything is positive $$ \displaystyle{\sqrt{\dfrac{a^2+ab+b^2}{3}}+\sqrt{ab}\leq a+b} $$$$\iff \frac{a^2+ab+b^2}{3} + ab + 2 \sqrt{\frac{ab(a^2 + ab + b^2)}{3}} \le a^2 + 2ab + b^2$$$$\iff 2 \sqrt{\frac{ab(a^2 + ab + b^2)}{3}} \le a^2 + ab + b^2 - \frac{a^2+ab+b^2}{3} = \frac{2}{3}(a^2 + ab + b^2)$$$$\iff \frac{a^3b + a^2b^2 + ab^3}{3} \le \frac{a^4 + b^4 + a^2b^2 + 2a^3b + 2ab^3 + 2a^2b^2}{9}$$$$\iff a^4 + b^4 + 3a^2b^2 + 2a^3b + 2ab^3 \ge 3a^3b + 3ab^3 + 3a^2b^2$$$$\iff a^4 + b^4 \ge a^3b + ab^3$$which is just rearrangement, and equality occurs iff $a=b$.

Let $ a$ and $ b$ are positive numbers . Prove that$$(\sqrt {a}+\sqrt {b})\sqrt {a+b}\geq\sqrt {a^2+b^2}$$$$a\sqrt {\frac{a}{a+2b}}+b\sqrt {\frac{b}{b+2a}}\geq\sqrt {\frac{2(a^2+b^2)}{3}}$$

Let $ a$ and $ b$ are positive numbers . Prove that $$\frac{a^2}{5a+3b}+\frac{b^2}{5b+3a}\geq\frac{\sqrt {2(a^2+b^2)}}{8}$$Let $ a,b$ and $ c$ are positive numbers . Prove that $$\frac{a^2}{a+k(b+c)}+\frac{b^2}{b+k(c+a)}+\frac{c^2}{c+k(a+b)}\geq\frac{\sqrt {3(a^2+b^2+c^2)}}{2k+1}$$Where $0<k\leq \frac{3}{11} $ or $k\geq \frac{6}{11}.$ (Lijvzhi)

sqing wrote: Let $ a$ and $ b$ are positive numbers . Prove that$$(\sqrt {a}+\sqrt {b})\sqrt {a+b}\geq\sqrt {a^2+b^2}$$ In fact, $$2\sqrt{a^2+b^2} \geq (\sqrt{a} + \sqrt{b})\sqrt{a+b} \geq \sqrt{a^2+b^2} $$$\bullet$ First part: $$2\sqrt{a^2+b^2} \geq (\sqrt{a} + \sqrt{b})\sqrt{a+b} \iff 4(a^2+b^2) \geq (\sqrt{a}+\sqrt{b})^2(a+b) \iff (3a+3b+4\sqrt{ab})(a+b -2\sqrt{ab}) \geq 0\blacksquare$$$\bullet$ Second part: $$(\sqrt{a} + \sqrt{b})\sqrt{a+b} \geq \sqrt{a^2+b^2} \iff (a+b)(\sqrt{a} + \sqrt{b})^2 \geq (a^2+b^2) \iff 2ab + 2(a+b)\sqrt{ab} \geq 0\blacksquare$$

Let $ a$ and $ b$ be non negative real numbers. Then $$2\sqrt{a^2+b^2} \geq (\sqrt{a} + \sqrt{b})\sqrt{a+b}\geq \sqrt{a^2+b^2} $$