a.

\(\dfrac{x}{x+\sqrt{3x+yz}}=\dfrac{x}{x+\sqrt{x\left(x+y+z\right)+yz}}=\dfrac{x}{x+\sqrt{\left(x+y\right)\left(z+x\right)}}\le\dfrac{x}{x+\sqrt{\left(\sqrt{xz}+\sqrt{xy}\right)^2}}\)

\(\Rightarrow\dfrac{x}{x+\sqrt{3x+yz}}\le\dfrac{x}{x+\sqrt{xy}+\sqrt{xz}}=\dfrac{\sqrt{x}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}\)

Tương tự:

\(\dfrac{y}{y+\sqrt{3y+xz}}\le\dfrac{\sqrt{y}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}\) ; \(\dfrac{z}{z+\sqrt{3z+xy}}\le\dfrac{\sqrt{z}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}\)

Cộng vế:

\(VT\le\dfrac{\sqrt{x}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}+\dfrac{\sqrt{y}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}+\dfrac{\sqrt{z}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}=1\) (đpcm)

Dấu "=" xảy ra khi \(x=y=z=1\)

b.

\(VP=\dfrac{4\left(a+b+c\right)}{2\sqrt{4a\left(a+3b\right)}+2\sqrt{4b\left(b+3c\right)}+2\sqrt{4c\left(c+3a\right)}}\)

\(VP\ge\dfrac{4\left(a+b+c\right)}{4a+a+3b+4b+b+3c+4c+c+3a}\)

\(VP\ge\dfrac{4\left(a+b+c\right)}{8\left(a+b+c\right)}=\dfrac{1}{2}\) (đpcm)

Dấu "=" xảy ra khi \(a=b=c\)

Áp dụng bđt Cô-si vào 2 số dương có:

\(\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{2}{\sqrt{xy}}\Rightarrow\dfrac{1}{2}\ge\dfrac{2}{\sqrt{xy}}\Rightarrow\sqrt{xy}\ge4\)

\(\Rightarrow\sqrt{x}+\sqrt{y}\ge2\sqrt{\sqrt{xy}}=2\sqrt{4}=4\)

Dấu = xảy ra \(\Leftrightarrow x=y=4\)

`1/x+1/y>=2/(\sqrt{xy})`

`<=>1/2>=2/(\sqrt{xy})`

`<=>\sqrt{xy}>=4`

`=>\sqrt{x}+\sqrt{y}>=2.2=4`

Dấu "=" xảy ra khi `x=y=4`

+\(10=x+3y=x+\frac{y}{3}+\frac{y}{3}+\frac{y}{3}+\frac{y}{3}+\frac{y}{3}+\frac{y}{3}+\frac{y}{3}+\frac{y}{3}+\frac{y}{3}\ge10\sqrt[10]{\frac{1}{3^9}x.y^9}\)

\(=\frac{10}{3}.\sqrt[10]{3}.\sqrt[10]{xy^9}\)

\(\Rightarrow xy^9\le3^9\)

+\(\frac{1}{\sqrt{x}}+\frac{27}{\sqrt{3y}}=\frac{1}{\sqrt{x}}+\frac{3}{\sqrt{3y}}+\frac{3}{\sqrt{3y}}+.....+\frac{3}{\sqrt{3y}}\)

\(\ge10\sqrt[10]{\frac{3^9}{\sqrt{3^9x.y^9}}}\ge10\sqrt[10]{\frac{3^9}{\sqrt{3^9.3^9}}}=10\)

Dấu "=" xảy ra khi và chỉ khi \(x=1;y=3\)

x + 25 = 64

x         = 64 - 25

x         = 39

Vậy x = 39

Áp dụng bất đẳng thức AM-GM cho ba số dương, ta có:

\(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{x}}+x\ge3\sqrt[3]{\frac{1}{\sqrt{x}}.\frac{1}{\sqrt{x}}.x}=3\left(1\right)\)

\(\frac{27}{\sqrt{3y}}+\frac{27}{\sqrt{3y}}+3y\ge3\sqrt[3]{\frac{27}{\sqrt{3y}}.\frac{27}{\sqrt{3y}}.3y}=27\left(2\right)\)

Cộng theo vế các bất đẳng thức (1) và (2) ta được: \(2\left(\frac{1}{\sqrt{x}}+\frac{27}{\sqrt{3y}}\right)+x+3y\ge30\)

\(\Rightarrow2\left(\frac{1}{\sqrt{x}}+\frac{27}{\sqrt{3y}}\right)\ge30-\left(x+3y\right)\ge20\)(Do theo giả thiết thì \(x+3y\le10\))

\(\Rightarrow\frac{1}{\sqrt{x}}+\frac{27}{\sqrt{3y}}\ge10\)(đpcm)

Đẳng thức xảy ra khi \(\hept{\begin{cases}x=1\\y=3\end{cases}}\)

CHÚC BẠN HỌC TỐT

\(\sqrt{x+yz}=\sqrt{x\left(x+y+z\right)+yz}=\sqrt{\left(x+y\right)\left(z+x\right)}\ge\sqrt{\left(\sqrt{xz}+\sqrt{xy}\right)^2}=\sqrt{xy}+\sqrt{xz}\)

\(\Rightarrow\dfrac{x}{x+\sqrt{x+yz}}\le\dfrac{x}{x+\sqrt{xy}+\sqrt{xz}}=\dfrac{\sqrt{x}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}\)

Tương tự:

\(\dfrac{y}{y+\sqrt{y+xz}}\le\dfrac{\sqrt{y}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}\)

\(\dfrac{z}{z+\sqrt{z+xy}}\le\dfrac{\sqrt{z}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}\)

Cộng vế:

\(VT\le\dfrac{\sqrt{x}+\sqrt{y}+\sqrt{z}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}=1\)

Dấu "=" xảy ra khi \(x=y=z=\dfrac{1}{3}\)

Lời giải:

Áp dụng BĐT SVac-xơ:

\(\frac{1}{\sqrt{x}}+\frac{27}{\sqrt{3y}}=\frac{1}{\sqrt{x}}+\frac{9}{\sqrt{3y}}+\frac{9}{\sqrt{3y}}+\frac{9}{\sqrt{3y}}\geq \frac{(1+3+3+3)^2}{\sqrt{x}+3\sqrt{3y}}\)

\(\Leftrightarrow \frac{1}{\sqrt{x}}+\frac{27}{\sqrt{3y}}\geq \frac{100}{x+3\sqrt{3y}}(1)\)

Áp dụng BĐT Bunhiacopxky:

\((x+3y)(1+9)\geq (\sqrt{x}+3\sqrt{3y})^2\)

\(\Rightarrow \sqrt{x}+3\sqrt{3y}\leq \sqrt{10(x+3y)}\leq 10(2)\) do \(x+3y\leq 10\)

Từ \((1);(2)\Rightarrow \frac{1}{\sqrt{x}}+\frac{27}{\sqrt{3y}}\geq \frac{100}{x+3\sqrt{3y}}\geq \frac{100}{10}=10\) (đpcm)

Dấu bằng xảy ra khi \(\frac{\sqrt{x}}{1}=\frac{\sqrt{3y}}{3}; x+3y=10\Rightarrow x=1;y=3\)

\(GT\Leftrightarrow xy=2\left(x+y\right)\ge4\sqrt{xy}\Rightarrow\sqrt{xy}\ge4\)

\(\Rightarrow4\le\sqrt{xy}\le\dfrac{1}{4}\left(\sqrt{x}+\sqrt{y}\right)^2\)

\(\Rightarrow\sqrt{x}+\sqrt{y}\ge4\)

Dấu "=" xảy ra khi \(x=y=4\)

Áp dụng BĐT cô si với ba số không âm ta có :

$\frac{1}{{\left(x+1\right)}^2}+\frac{x+1}{8}+\frac{x+1}{8}\ge 3\sqrt[3]{3\frac{1}{64}}=\frac{3}{4}$

=> $\frac{1}{{\left(x+1\right)}^2}\ge \frac{3}{4}-\frac{x+1}{4}$ (1)

Dấu '' = '' xảy ra khi x = 1 

CM tương tự ra có " $\frac{1}{{\left(y+1\right)}^2}\ge \frac{3}{4}-\frac{y+1}{4}$(2) ; $\frac{1}{{\left(z+1\right)}^2}\ge \frac{3}{4}-\frac{z+1}{4}$ (3)

Dấu ''= '' xảy ra khi y = 1 ; z = 1 

Từ (1) (2) và (3) => $\frac{1}{{\left(x+1\right)}^2}+\frac{1}{{\left(y+1\right)}^2}+\frac{1}{{\left(z+1\right)}^2}\ge \frac{3}{4}\cdot 3-\frac{x+y+z+3}{4}$$\ge \frac{9}{4}-\frac{3\sqrt[3]{3xyz}+3}{4}=\frac{9}{4}-\frac{6}{4}=\frac{3}{4}$

BĐT được chứng minh 

Dấu '' = '' của bất đẳng thức xảy ra khi x =y =z = 1

:()