$1\leq\frac{(x+y)(x^{3}+y^{3})}{(x^{2}+y^{2})^{2}}\leq\frac{9}8 $ Indeed, the left inequality is obvious by Cauchy. We'll proceed now to prove the right one: $ \frac{(x+y)(x^{3}+y^{3})}{(x^{2}+y^{2})^{2}}\leq\frac{9}8 $ becomes, after some computations, $ x^4+y^4+18x^2y^2 \leq 8xy(x^2+y^2) \Leftrightarrow \frac{x^2}{y^2}+\frac{y^2}{x^2}+18 \ge 8(\frac{x}{y}+\frac{y}{x} )$ (*). Denoting $\frac{x}{y}=a$, it is true that (*) is equivalent with $ a^4-8a^3+18a^2-8a+1 \ge 0$, or $(a^2-4a+1)^2 \geq 0 $.

first we have $(x^3+y^3).(x+y)>=(x^2+y^2)^2$ so $(x^3+y^3).(x+y)/(x^2+y^2)^2>=1$ for part two we should proof: $8(x^4+y^4+xy^3+yx^3)=<9(x^4+y^4+2x^2y^2)$ $8xy^3+8yx^3=<x^4+y^4+18x^2y^2$ $(4xy^3+4yx^3-8x^2y^2)=<(x^4+y^4+6x^2y^2-4xy^3-4yx^3)+4x^2y^2$ $4xy(x-y)^2=<(x-y)^4+4x^2y^2$ $0=<((x-y)^2-2xy)^2$

part 1 denote by x+y=s, xy=p $ \Rightarrow 1\leq \frac{s(s^{3}-3ps)}{(s^{2}-2p)^{2}}\Leftrightarrow ps^{2}-4p^{2}\geq 0\Leftrightarrow p(s^{2}-4p)\geq 0 $ part 2: $ \frac{s(s^{3}-3ps)}{(s^{2}-2p)^{2}}\leq \frac{9}{8}\Leftrightarrow s^{4}-12ps^{2}+36p^{2}\geq 0\Leftrightarrow (s^{2}-6p)^{2}\geq 0 $

The left side is obvious from the Cauchy-Schwarz inequality. For the right side, we can rewrite it into \[(x+y)^2(x^2-xy+y^2)\leq \frac 98(x^2+y^2)^2.\] Note that we have, using the AM-GM inequality, \[\begin{aligned}8(x+y)^2(x^2-xy+y^2)&=4(x^2+2xy+y^2)(2x^2-2xy+2y^2)\\&\leq \left(3x^2+3y^2\right)^2\\&=9(x^2+y^2)^2.\end{aligned}\] Hence we are done. Equality occurs if and only if $(x,y)=\Bigl((\sqrt3+1)t,(\sqrt3-1)t\Bigr).$ $\Box$

hello, we get $\frac{9}{8}-\frac{(x+y)(x^3+y^3)}{(x^2+y^2)^2}=\frac{1}{8}\frac{(y^2-4xy+x^2)^2}{(x^2+y^2)^2}\geq 0$ Sonnhard.

hello, and $\frac{(x+y)(x^3+y^3)}{(x^2+y^2)^2}-1=\frac{xy(x-y)^2}{(x^2+y^2)^2}\geq 0$. Sonnhard.

Part 1: $\frac{(x+y)(x^3+y^3)}{(x^2+y^2)^2} \ge 1$ By the Cauchy–Schwarz inequality, $ (\sqrt{x}^2+\sqrt{y}^2)(\sqrt{x^3}^2+\sqrt{x^3}^2) \ge (\sqrt{x}\sqrt{x^3}+\sqrt{y}\sqrt{y^3})^2 = (x^2+y^2)^2$. Part 2: $\frac{(x+y)(x^3+y^3)}{(x^2+y^2)^2} \le \frac{9}{8}$ We have to prove that $8(x+y)(x^3+y^3) \le 9(x^2+y^2)^2 \iff 8(x^4+x^3y+xy^3+y^4) \le 9(x^4+2x^2y^2+y^4)$ $ \iff 8xy(x^2+y^2) \le (x^2+y^2)^2+(4xy)^2$ which is true by AM-GM inequality.

drEdrE wrote: $1\leq\frac{(x+y)(x^{3}+y^{3})}{(x^{2}+y^{2})^{2}}\leq\frac{9}8 $ Indeed, the left inequality is obvious by Cauchy. We'll proceed now to prove the right one: $ \frac{(x+y)(x^{3}+y^{3})}{(x^{2}+y^{2})^{2}}\leq\frac{9}8 $ becomes, after some computations, $ x^4+y^4+18x^2y^2 \leq 8xy(x^2+y^2) \Leftrightarrow \frac{x^2}{y^2}+\frac{y^2}{x^2}+18 \ge 8(\frac{x}{y}+\frac{y}{x} )$ (*). Denoting $\frac{x}{y}=a$, it is true that (*) is equivalent with $ a^4-8a^3+18a^2-8a+1 \ge 0$, or $(a^2-4a+1)^2 \geq 0 $. Nice solution

WLOG, $x+y=1$ and $t=xy$. Then we need to show that $$1\le\frac{1-3t}{(2t-1)^2}\le\frac98.$$Let $f(t)$ be this expression, which is continuously differentiable. We have $1=x+y\ge2\sqrt{xy}$, hence $t\le\frac14$. We have: $$f'(t)=\frac{6t-1}{(2t-1)^3},$$so we just need to check $t=0,t=\frac16$, and $t=\frac14$, all of which satisfy the inequality.

Let $x,y $ be reals . Show that $$ (x+y)(x^3+y^3)\leq \frac98(x^2+y^2)^2$$

sqing wrote: Let $x,y $ be reals . Show that $$ (x+y)(x^3+y^3)\leq \frac98(x^2+y^2)^2$$ Open L.H.S and R.H.S

Thanks.

Let $x,y $ be reals such that $x^2+y^2\neq 0 . $ Show that $$-1\leq \frac{(x+y)(x^3+y^3-2xy(x+y))}{(x^2+y^2)^2}\leq \frac{25}{24}$$$$-2\leq \frac{(x+y)(x^3+y^3-3xy(x+y))}{(x^2+y^2)^2}\leq \frac98$$

sqing wrote: Let $x,y $ be reals . Show that $$ (x+y)(x^3+y^3)\leq \frac98(x^2+y^2)^2$$ sqing wrote: Let $x,y $ be reals such that $x^2+y^2\neq 0 . $ Show that $$-1\leq \frac{(x+y)(x^3+y^3-2xy(x+y))}{(x^2+y^2)^2}\leq \frac{25}{24}$$$$-2\leq \frac{(x+y)(x^3+y^3-3xy(x+y))}{(x^2+y^2)^2}\leq \frac98$$

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