1) Áp dụng bđt Cauchy cho 3 số dương ta có

 \(\dfrac{1}{x}+\dfrac{1}{x}+\dfrac{1}{x}+x^3\ge4\sqrt[4]{\dfrac{1}{x}.\dfrac{1}{x}.\dfrac{1}{x}.x^3}=4\) (1)

\(\dfrac{3}{y^2}+y^2\ge2\sqrt{\dfrac{3}{y^2}.y^2}=2\sqrt{3}\) (2)

\(\dfrac{3}{z^3}+z=\dfrac{3}{z^3}+\dfrac{z}{3}+\dfrac{z}{3}+\dfrac{z}{3}\ge4\sqrt[4]{\dfrac{3}{z^3}.\dfrac{z}{3}.\dfrac{z}{3}.\dfrac{z}{3}}=4\sqrt{3}\) (3)

Cộng (1);(2);(3) theo vế ta được

\(\left(\dfrac{3}{x}+\dfrac{3}{y^2}+\dfrac{3}{z^3}\right)+\left(x^3+y^2+z\right)\ge4+2\sqrt{3}+4\sqrt{3}\)

\(\Leftrightarrow3\left(\dfrac{1}{x}+\dfrac{1}{y^2}+\dfrac{1}{z^3}\right)\ge3+4\sqrt{3}\)

\(\Leftrightarrow P\ge\dfrac{3+4\sqrt{3}}{3}\)

Dấu "=" xảy ra <=> \(\left\{{}\begin{matrix}\dfrac{1}{x}=x^3\\\dfrac{3}{y^2}=y^2\\\dfrac{3}{z^3}=\dfrac{z}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=\sqrt[4]{3}\\z=\sqrt{3}\end{matrix}\right.\) (thỏa mãn giả thiết ban đầu)

2) Ta có \(4\sqrt{ab}=2.\sqrt{a}.2\sqrt{b}\le a+4b\)

Dấu"=" khi a = 4b

nên \(\dfrac{8}{7a+4b+4\sqrt{ab}}\ge\dfrac{8}{7a+4b+a+4b}=\dfrac{1}{a+b}\)

Khi đó \(P\ge\dfrac{1}{a+b}-\dfrac{1}{\sqrt{a+b}}+\sqrt{a+b}\)

Đặt \(\sqrt{a+b}=t>0\) ta được

\(P\ge\dfrac{1}{t^2}-\dfrac{1}{t}+t=\left(\dfrac{1}{t^2}-\dfrac{2}{t}+1\right)+\dfrac{1}{t}+t-1\)

\(=\left(\dfrac{1}{t}-1\right)^2+\dfrac{1}{t}+t-1\)

Có \(\dfrac{1}{t}+t\ge2\sqrt{\dfrac{1}{t}.t}=2\) (BĐT Cauchy cho 2 số dương)

nên \(P=\left(\dfrac{1}{t}-1\right)^2+\dfrac{1}{t}+t-1\ge\left(\dfrac{1}{t}-1\right)^2+1\ge1\)

Dấu "=" xảy ra <=> \(\left\{{}\begin{matrix}\dfrac{1}{t}-1=0\\t=\dfrac{1}{t}\end{matrix}\right.\Leftrightarrow t=1\)(tm)

khi đó a + b = 1

mà a = 4b nên \(a=\dfrac{4}{5};b=\dfrac{1}{5}\)

Vậy MinP = 1 khi \(a=\dfrac{4}{5};b=\dfrac{1}{5}\)

ÁP dụng BĐT Mincopxki, ta có:

\(A\ge\sqrt{\left(x+y\right)^2+\left(\dfrac{1}{x}+\dfrac{1}{y}\right)^2}\)

\(=\sqrt{\left(x+y\right)^2+\dfrac{\left(x+y\right)^2}{\left(xy\right)^2}}\)

\(\ge\sqrt{2\sqrt{\left(x+y\right)^2.\dfrac{\left(x+y\right)^2}{\left(xy\right)^2}}}=\sqrt{\dfrac{2\left(x+y\right)^2}{xy}}\) (cô si)

\(\ge\sqrt{\dfrac{2.4xy}{xy}}=\sqrt{8}=2\sqrt{2}\left(Côsi\right)\)

Min \(A=2\sqrt{2}\Leftrightarrow x=y\)

Đặt \(\left\{{}\begin{matrix}\sqrt{y+z-4}=a>0\\\sqrt{z+x-4}=b>0\\\sqrt{x+y-4}=c>0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{b^2+c^2-a^2+4}{2}\\y=\dfrac{c^2+a^2-b^2+4}{2}\\z=\dfrac{a^2+b^2-c^2+4}{2}\end{matrix}\right.\).

\(2P=\dfrac{b^2+c^2-a^2+4}{a}+\dfrac{c^2+a^2-b^2+4}{b}+\dfrac{a^2+b^2-c^2+4}{c}=\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{a}+\dfrac{b^2}{a}+\dfrac{c^2}{b}+\dfrac{a^2}{c}+\dfrac{4}{a}+\dfrac{4}{b}+\dfrac{4}{c}-a-b-c\).

Áp dụng bất đẳng thức AM - GM:

\(\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{a}=\left(\dfrac{a^2}{b}+b\right)+\left(\dfrac{b^2}{c}+c\right)+\left(\dfrac{c^2}{a}+a\right)-\left(a+b+c\right)\ge2a+2b+2c-a-b-c=a+b+c\).

Tương tự, \(\dfrac{b^2}{a}+\dfrac{c^2}{b}+\dfrac{a^2}{c}\ge a+b+c\).

Do đó \(2P\ge a+b+c+\dfrac{4}{a}+\dfrac{4}{b}+\dfrac{4}{c}=\left(a+\dfrac{4}{a}\right)+\left(b+\dfrac{4}{b}\right)+\left(c+\dfrac{4}{c}\right)\ge4+4+4=12\Rightarrow P\ge6\).

Đẳng thức xảy ra khi a = b = c = 2 hay x = y = z = 4.

Vậy Min P = 6 khi x = y = z = 4.

\(P=\dfrac{4x}{2.2.\sqrt{y+z-4}}+\dfrac{4y}{2.2.\sqrt{x+z-4}}+\dfrac{4z}{2.2.\sqrt{x+y-4}}\)

\(P\ge4\left(\dfrac{x}{y+z}+\dfrac{y}{z+x}+\dfrac{z}{x+y}\right)\ge4.\dfrac{3}{2}=6\)

Dấu "=" xảy ra khi \(x=y=z=4\)

đặt \(\sqrt{y}=b;x=a\)

Ta có : \(\sqrt{x+1}\) có nghĩa khi `x >= -1`  Từ đk ta có :

\(x+2\left(1+\sqrt{x+1}\right)=x+1+2\sqrt{x+1}+1=\left(\sqrt{x+1}+1\right)^2\)

\(\Leftrightarrow\sqrt{x+2\left(1+\sqrt{x+1}\right)}=\sqrt{x+1}+1\)

\(x+2\left(1-\sqrt{x+1}\right)=x+1-2\sqrt{x+1}+1=\left(\sqrt{x+1}-1\right)^2\\ \Leftrightarrow\sqrt{x+2\left(1-\sqrt{x+1}\right)}=\left|\sqrt{x+1}-1\right|\)

Ta có : \(y=\sqrt{x+1}+1+\left|\sqrt{x+1}-1\right|\)  `(1)`

Ta bỏ dấu \(\left|\right|\) ở `1`

Ta có TH :

`-1<= x <= 0` ; lúc này \(\sqrt{x+1}-1\le0\)

nên : \(\left|\sqrt{x+1}-4\right|=1-\sqrt{x+1}\)

`1` trở thành : `y=2`

`x>0` lúc này \(\sqrt{x+1}-1>0\) nên

\(\left|\sqrt{x+1}-1\right|=\sqrt{x+1}-1\)

`1` trở thành : \(y=2\sqrt{x+1}>2\left(x>0\right)\)

Vì : \(y=\left\{{}\begin{matrix}2khi-1\le x\le0\\2\sqrt{x+1}kh\text{i}>0\end{matrix}\right.\)

gtnn của `y=2` với mọi \(x\in\left[-1;0\right]\)

Lời giải:

Áp dụng BĐT Mincopxky:

\(y=\sqrt{x^2+4x+8}+\sqrt{x^2-4x+8}=\sqrt{(x+2)^2+4}+\sqrt{(x-2)^2+4}\)

\(=\sqrt{(x+2)^2+2^2}+\sqrt{(2-x)^2+2^2}\geq \sqrt{(x+2+2-x)^2+(2+2)^2}\)

\(=\sqrt{32}=4\sqrt{2}\)

Vậy $y_{\min}=4\sqrt{2}$ khi $x=0$

\(x=\dfrac{1}{\sqrt{2}}\left(\sqrt{4+2\sqrt{3}}+\sqrt{4-2\sqrt{3}}\right)\)

\(=\dfrac{1}{\sqrt{2}}\left(\sqrt{\left(\sqrt{3}+1\right)^2}+\sqrt{\left(\sqrt{3}-1\right)^2}\right)=\sqrt{6}\)

\(y=\sqrt{\left(\sqrt{6}-1\right)^2}=\sqrt{6}-1\)

\(\Rightarrow x-y=1\Rightarrow P=1\)

\(B=x-2020-\sqrt{x-2020}+\dfrac{1}{4}+\dfrac{8079}{4}\)

\(B=\left(\sqrt{x-2020}-\dfrac{1}{2}\right)^2+\dfrac{8079}{4}\ge\dfrac{8079}{4}\)

\(B_{min}=\dfrac{8079}{4}\) khi \(x=\dfrac{8081}{4}\)

Lời giải:
TXĐ: $[-1;1]$

$y'=\frac{1}{2\sqrt{x+1}}-\frac{1}{2\sqrt{1-x}}+\frac{x}{2}$

$y'=0\Leftrightarrow x=0$

$f(0)=2$;

$f(1)=f(-1)=\sqrt{2}+\frac{1}{4}$
Vậy $f_{\min}=2; f_{\max}=\frac{1}{4}+\sqrt{2}$

Bạn coi lại đề, nhìn 2 vế của điều kiên đều là \(\sqrt{x+2}\) có vẻ sai sai rồi đó