Here is what I did : First if $x+y \leq 0$ the result is obvious, since the Lhs is neagtive and the Rhs is positive. Then, if $x+y > 0$ but $y < 0$, just remark that $x+y \leq x + |y|$ and $x^2-xy+y^2 \geq x^2 - x|y| + y^2 > 0$ so that the LHs is smaller than if we replace $y$ by $|y|$ but the Rhs remains the same we this operation. Thus, we may assume that $x,y$ are nonnegative. The desired inequality is equivalent to $ (x+y)^2 \sqrt {x^2+y^2} \leq 2 \sqrt {2} (x^3 + y^3)$. This folows easily from the inequalities between means of order 1 and 3, and of order 2 and 3, that is : $ \frac {x+y} {2} \leq \frac {x^3+y^3} {2}$ and $ \sqrt {\frac {x^2+y^2} {2} } \leq (\frac {x^3+y^3} {2})^ {\frac 1 3}$. Pierre.

Or just do this: $x+y>0$ so \begin{eqnarray*} x+y &\leq& 2\sqrt{\frac{x^2+y^2}{2}} \\ x^2+2xy+y^2 & \leq & 2 x^2+2y^2 \\ 0 &\leq & (x-y)^2 \\ \frac{\sqrt{2} \cdot \sqrt{x^2+y^2}}{x^2-xy+y^2} &\leq & \frac{2\sqrt{2}}{\sqrt{x^2+y^2}} \\ x^2+y^2 &\leq& 2x^2-2xy+y^2 \\ 0 &\leq& (x-y)^2 \end{eqnarray*}

My solution of problem 1: $xy \leq \frac{x^2+y^2}{2}$, therefore $x^2-xy+y^2 \geq \frac{x^2+y^2}{2}$ Cauchy tells us that $\left(x+y\right)^2 \leq 2\left(x^2+y^2\right)$ thus $x+y \leq \sqrt{2\left(x^2+y^2\right)}$ Applying the two above conclusions we get the result we are looking for.

Is it legitimate to convert to polar coordinates to do an inequality? Cause that, mixed with a little calculus works pretty well for this.

1. If (x,y) is a point in the plane, write it in polar coordinates, so x = r cos t, y = r sin t. We start with sin (t+pi/4) + sin (2t) <= 2 (because sin is less than 1) -> sin (t+pi/4) / (2 - sin (2t)) <= 1 (we can divide by 2-sin(2t)) -> (sin t/sqrt(2) + cos t/sqrt(2)) / (2 - 2 sin t cos t) <= 1 (expand) -> (r/r^2)*((sin t + cos t)/(1 - sin t cos t)) <= 2sqrt(2)/r (algebra) -> (x+y)/(x^2-xy+y^2) <= 2sqrt(2)/sqrt(x^2+y^2) (writing in Cartesian coordinates). I agree the other solution is shorter; I just found this interesting.

You can also prove the inequality by squaring it (in fact, the right hand side of the inequality is obviously $\geq 0$; if the left hand side is $\leq 0$, then the inequality is trivial, so it is enough to consider the case when it is $\geq 0$ as well, and then we can square the inequality); this leads to $\frac{\left( x+y\right) ^2}{\left( x^2-xy+y^2\right) ^2}\leq \frac{8}{x^2+y^2}$. This is obviously equivalent to $\left( x+y\right) ^2\left( x^2+y^2\right) \leq 8\left( x^2-xy+y^2\right) ^2$. But actually, an easy calculation shows that $8\left( x^2-xy+y^2\right) ^2-\left( x+y\right) ^2\left( x^2+y^2\right) =\left( x-y\right) ^2\left( 7x^2-4xy+7y^2\right)$ $=\left( x-y\right) ^2\left( 2\left( x-y\right) ^2+5x^2+5y^2\right) \geq 0$, so everything is proven. Darij

If we suppose without loss of generality that $ x\geq y > 0$, and put $ x/y = t$, $ t\geq1$ then we can also solve the inequality. This way works for many hommogenous inequalities with two variables. Let's check it.

Expanding, it's equivalent to $ 7x^4 + 22x^2y^2 + 7y^4 \ge 18x^3y + 18xy^3$. But $ 7(x - y)^4 \ge 0 \implies 7x^4 + 7y^4 + 42x^2y^2 \ge 28x^3y + 28xy^3$, so using this we have to show $ 10x^3y + 10xy^3 \ge 20x^2y^2$, which is obvious by AM-GM.

I rearranged terms to get that we needed to prove that $\frac{x+y}{2}*\sqrt{\frac{x^2+y^2}{2}}\leq x^2-xy+y^2$. It's easy to see that $\frac{x+y}{2}\leq \sqrt{\frac{x^2+y^2}{2}}$, so $\frac{x+y}{2}*\sqrt{\frac{x^2+y^2}{2}}\leq \left( \sqrt{\frac{x^2+y^2}{2}}\right)^2=\frac{x^2+y^2}{2}$, so it suffices to show that $\frac{x^2+y^2}{2}\leq x^2-xy+y^2$. Moving all the terms to the Right-hand-side gives $0\leq \frac{(x-y)^2}{2}$, which is true. We can then work backwards to the given inequality.

by Caushi aa-ab+bb is bigger or egual to aa+bb/2 that means we must show aa+bb bigger or egual to 2ab.That is already true.

pbornsztein wrote: Prove that the inequality \[ \frac{ x+y}{x^2-xy+y^2 } \leq \frac{ 2\sqrt 2 }{\sqrt{ x^2 +y^2 } } \] holds for all real numbers $x$ and $y$, not both equal to 0. $(x+y)^2(x^2+y^2) \leq 8(x^4+y^4+x^2y^2 + 2x^2y^2 - 2x^3y - 2xy^3)$ or $x^4+y^4 + 2x^2y^2 + 10(x^3y+xy^3) \leq 8x^4 + 8y^4 + 24x^2y^2$ and we get $7x^4 + 7y^4 + 22x^2y^2 \ge 10(x^3y+xy^3)$ but $x^4 + x^2y^2 \ge 2x^3y$ (AM-GM) and we are done.

Since the inequality is homogeneous, we can assume WLOG that xy = 1. Now, substituting $m = (x+y)^2$, we have: $m = x^2 + y^2 + 2xy = x^2 + y^2 + 2 $ $ \geq 2\sqrt {xy} + 2 = 4$, thus we have $m \geq 4 $ Now squaring both sides of the inequality, we get: $$ \frac{m}{(m-3)^2 } \leq \frac{8}{m-2} $$after cross multiplication and simplification we get: $7m^2 -46m + 72 \geq 0$ or, $7(m-4)^2 +10(m-4) \geq 0$ which is always true since $m \geq 4$.

Let $x$ and $y$ be real numbers , not both equal to 0. Prove that $$\frac{(x+y)\sqrt{ x^2 +xy+y^2 }}{x^2-xy+y^2 } \leq 2\sqrt 3$$$$\frac{(x+y)\sqrt{ x^2 -xy+y^2 }}{x^2+xy+y^2 } \leq \frac{3}{2\sqrt 2}$$$$\frac{(x+y)\sqrt{ 2x^2 -xy+2y^2 }}{x^2+xy+y^2 } \leq \frac{5}{2\sqrt 3}$$$$\frac{(x+y)\sqrt{ 3x^2 -xy+3y^2 }}{x^2+xy+y^2 } \leq \frac{7}{4}$$$$ \frac{(x+y)\sqrt{ 2x^2 -xy+y^2 }}{x^2+xy+y^2 } <\frac{3}{2}$$$$\frac{(x+2y)\sqrt{ 2x^2 -xy+y^2 }}{x^2+xy+y^2 }<3$$

Let $a,b\ge0: a+b>0$. Prove that: $$\frac{1}{a^2+b^2}+\frac{2}{(a+b)^2}\ge\frac{8\sqrt{ab}}{(a+b)^3}$$

jokehim wrote: Let $a,b\ge0: a+b>0$. Prove that: $$\frac{1}{a^2+b^2}+\frac{2}{(a+b)^2}\ge\frac{8\sqrt{ab}}{(a+b)^3}$$ https://artofproblemsolving.com/community/c4h2681816p23257285 DAVROS wrote: Homogeneous so let $a+b=1$ and it becomes $\frac{1}{1-2ab}+2 \ge 8\sqrt{ab} \iff 16x^3-4x^2-8x+3\ge 0$ where $x=\sqrt{ab}$ Factor to $(2x-1)^2(4x+3)\ge 0$ which is true. dragonheart6 wrote: I saw the following problem today. I can not find it now. I give a proof here. Problem: Let $a, b > 0$. Prove that $$\frac{1}{a^4 + b^4} + 4(a + b) \ge \frac{5}{2} + 6\sqrt{ab}.$$ My proof: We have $$\frac{1}{a^4 + b^4} \ge \frac{1}{a^4 + b^4 + (a - b)^4} = \frac{1}{2(a^2 - ab + b^2)^2},$$and $$3(a + b) - 6\sqrt{ab} = 3(\sqrt{a} - \sqrt{b})^2 = \frac{3(a - b)^2}{(\sqrt{a} + \sqrt{b})^2} \ge \frac{3(a - b)^2}{2(a + b)}.$$ Thus, it suffices to prove that $$\frac{1}{2(a^2 - ab + b^2)^2} + a + b + \frac{3(a - b)^2}{2(a + b)} \ge \frac{5}{2}$$or $$\frac{1}{(a^2 - ab + b^2)^2} + \frac{2(a^2 - ab + b^2)}{a + b} + 3\cdot\frac{a^2 + b^2}{a + b} \ge 5.$$Using AM-GM, it suffices to prove that $$5\sqrt[5]{\frac{1}{(a^2 - ab + b^2)^2} \cdot \frac{2(a^2 - ab + b^2)}{a + b} \cdot \left(\frac{a^2 + b^2}{a + b}\right)^3} \ge 5$$or $$2(a^2 + b^2)^3 - (a^2 - ab + b^2)(a + b)^4 \ge 0$$or $$(a^2 + ab + b^2)(a - b)^4 \ge 0.$$ We are done.

Very nice, thanks.

jokehim wrote: For any non-negative real numbers: $a+b>0$. Prove that: $$\frac{a^3}{a^2+b^2}+b\ge\frac{a}{2}+\sqrt{ab}$$

For any non-negative real numbers: $a+b>0$. Prove that $$\frac{a^3}{a^2+b^2}+\frac{b}{2}\ge a$$ jokehim wrote: For any non-negative real numbers: $a+b>0$. Prove that: $$\frac{a^3}{a^2+b^2}+b\ge\frac{a}{2}+\sqrt{ab}$$ DAVROS wrote: True for $a=0$ so suppose $a>0$ and then because it is homogeneous you can set $a=1$ It becomes $\frac{1}{1+b^2} + b \ge \frac12+\sqrt{b}$ or $(x-1)^2 (2x^4+2x^3+x^2+1)\ge 0$ where $x=\sqrt b\ge0$

sqing wrote: Let $x$ and $y$ be real numbers , not both equal to 0. Prove that $$\frac{(x+y)\sqrt{ x^2 +xy+y^2 }}{x^2-xy+y^2 } \leq 2\sqrt 3$$$$\frac{(x+y)\sqrt{ x^2 -xy+y^2 }}{x^2+xy+y^2 } \leq \frac{3}{2\sqrt 2}$$$$\frac{(x+y)\sqrt{ 2x^2 -xy+2y^2 }}{x^2+xy+y^2 } \leq \frac{5}{2\sqrt 3}$$$$\frac{(x+y)\sqrt{ 3x^2 -xy+3y^2 }}{x^2+xy+y^2 } \leq \frac{7}{4}$$$$ \frac{(x+y)\sqrt{ 2x^2 -xy+y^2 }}{x^2+xy+y^2 } <\frac{3}{2}$$$$\frac{(x+2y)\sqrt{ 2x^2 -xy+y^2 }}{x^2+xy+y^2 }<3$$ h

Attachments:

Note that $x^2 + y^2 - xy \ge \frac{x^2+y^2}{2}$ so we need to prove $(x+y)\sqrt{x^2+y^2} \le \sqrt{2}{x^2+y^2}$ or $\frac{x+y}{\sqrt{2}} \le \sqrt{x^2 + y^2}$ or $x+y \le \sqrt{(1+1)(x^2 + y^2)}$ which is true by Cauchy.