/\(2020\left(\dfrac{1}{x^2+y^2}+\dfrac{1}{y^2+z^2}+\dfrac{1}{x^2+y^2}\right)ápdụngBDT\)

\(\dfrac{1}{x^2+y^2}+\dfrac{1}{y^2+z^2}+\dfrac{1}{x^2+z^2}\ge\dfrac{9}{2\left(x^2+y^2+z^2\right)}=\dfrac{9}{2\cdot2020}\)

\(ápdụngBĐTcosi\)

\(x^3+y^3+z^3\ge3xyz\)

\(\)=> VP\(\ge\) 9/2

Lời giải:Vì $x^2+y^2+z^2=2$ nên:

$P=\frac{x^2+y^2+z^2}{x^2+y^2}+\frac{x^2+y^2+z^2}{y^2+z^2}+\frac{x^2+y^2+z^2}{z^2+x^2}-\frac{x^3+y^3+z^3}{2xyz}$

$=3+\frac{x^2}{y^2+z^2}+\frac{y^2}{x^2+z^2}+\frac{z^2}{x^2+y^2}-\frac{x^3+y^3+z^3}{2xyz}$

$\leq 3+\frac{x^2}{2yz}+\frac{y^2}{2xz}+\frac{z^2}{2xy}-\frac{x^3+y^3+z^3}{2xyz}$

(theo BĐT AM-GM)

$=3+\frac{x^3+y^3+z^3}{2xyz}-\frac{x^3+y^3+z^3}{2xyz}=3$

Vậy $P_{\max}=3$

Dấu "=" xảy ra khi $x=y=z=\sqrt{\frac{2}{3}}$

Đề bài yêu cầu gì vậy em.

Đề lỗi công thức rồi. Bạn xem lại.

a/

$x^2+y^2+z^2=3$

$F=\dfrac{x^2+1}{z+2}+\dfrac{y^2+1}{x+2}+\dfrac{z^2+1}{y+2}$

$\ge \dfrac{(x+y+z)^2}{(x+y+z)+6}\qquad (\text{Titu})$

Đặt $t=x+y+z$.

Ta có $t^2\le 3(x^2+y^2+z^2)=9$

$\Rightarrow t\le 3$.

Xét $f(t)=\dfrac{t^2}{t+6}$ thì $f'(t)=\dfrac{t(t+12)}{(t+6)^2}>0$ nên $f(t)$ tăng trên $(0,+\infty)$.

Mặt khác $t\ge \sqrt{x^2+y^2+z^2}=\sqrt3$.

Suy ra $F\ge f(\sqrt3)=\dfrac{3}{6+\sqrt3}$ $=\dfrac{6-\sqrt3}{11}$.

Dấu bằng khi $x=y=z=1$.

$\boxed{\min F=\dfrac{6-\sqrt3}{11}}$.

b/

Đặt $S=\sqrt{\dfrac{a}{a+3}}+\sqrt{\dfrac{b}{b+3}}+\sqrt{\dfrac{c}{c+3}}.$

Theo bất đẳng thức Cauchy-Schwarz, $S^2\le (a+b+c)\left(\dfrac1{a+3}+\dfrac1{b+3}+\dfrac1{c+3}\right)$.

Lại có $(a+b+c)^2\ge 3(ab+bc+ca)=9$

$\Rightarrow a+b+c\ge 3$.

Theo bất đẳng thức Nesbitt dạng Engel,

$\dfrac1{a+3}+\dfrac1{b+3}+\dfrac1{c+3}\le \dfrac1{6}(3)=\dfrac12.$

Do đó $S^2\le \dfrac32$

$\Rightarrow S\le \sqrt{\dfrac32}<\dfrac32$.

Suy ra $\sqrt{\dfrac{a}{a+3}}+\sqrt{\dfrac{b}{b+3}}+\sqrt{\dfrac{c}{c+3}}\le \dfrac32.$

\(\frac{x^2}{y^2}+\frac{y^2}{z^2}\ge2\cdot\sqrt{\frac{x^2}{y^2}\cdot\frac{y^2}{z^2}}=2\cdot\sqrt{\frac{x^2}{z^2}}=2\cdot\frac{x}{z}\)

\(\frac{y^2}{z^2}+\frac{z^2}{x^2}\ge2\cdot\sqrt{\frac{y^2}{z^2}\cdot\frac{z^2}{x^2}}=2\cdot\sqrt{\frac{y^2}{x^2}}=2\cdot\frac{y}{x}\)

\(\frac{z^2}{x^2}+\frac{x^2}{y^2}\ge2\cdot\sqrt{\frac{z^2}{x^2}\cdot\frac{x^2}{y^2}}=2\cdot\sqrt{\frac{z^2}{y^2}}=2\cdot\frac{z}{y}\)

Do đó; \(\left(\frac{x^2}{y^2}+\frac{y^2}{z^2}\right)+\left(\frac{y^2}{z^2}+\frac{z^2}{x^2}\right)+\left(\frac{z^2}{x^2}+\frac{x^2}{y^2}\right)\ge2\cdot\frac{x}{z}+2\cdot\frac{y}{x}+2\cdot\frac{z}{y}\)

=>\(2\left(\frac{x^2}{y^2}+\frac{y^2}{z^2}+\frac{z^2}{x^2}\right)\ge2\left(\frac{x}{z}+\frac{y}{x}+\frac{z}{y}\right)\)

=>\(\frac{x^2}{y^2}+\frac{y^2}{z^2}+\frac{z^2}{x^2}\ge\frac{x}{z}+\frac{y}{x}+\frac{z}{y}\)