Mình áp dụng luôn Cô - si cho các số ta được

a) \(\frac{x}{2}+\frac{18}{x}\ge2\sqrt{\frac{x}{2}\cdot\frac{18}{x}}=2.\sqrt{9}=2.3=6\)

b) \(y=\frac{x}{2}+\frac{2}{x-1}=\frac{x-1}{2}+\frac{2}{x-1}+\frac{1}{2}\ge2\sqrt{\frac{x-1}{2}\cdot\frac{2}{x-1}}+\frac{1}{2}=2+\frac{1}{2}=\frac{5}{2}\)

c) \(\frac{3x}{2}+\frac{1}{x+1}=\frac{3\left(x+1\right)}{2}+\frac{1}{x+1}-\frac{3}{2}\ge2\sqrt{\frac{3\left(x+1\right)}{2}\cdot\frac{1}{x+1}}-\frac{3}{2}=2\sqrt{\frac{3}{2}}-\frac{3}{2}=\frac{-3+2\sqrt{6}}{2}\)

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

g) \(\frac{x^2+4x+4}{x}=\frac{\left(x+2\right)^2}{x}\ge0\)

1.

\(6=\frac{\sqrt{2}^2}{x}+\frac{\sqrt{3}^2}{y}\ge\frac{\left(\sqrt{2}+\sqrt{3}\right)^2}{x+y}=\frac{5+2\sqrt{6}}{x+y}\)

\(\Rightarrow x+y\ge\frac{5+2\sqrt{6}}{6}\)

Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}\frac{x}{\sqrt{2}}=\frac{y}{\sqrt{3}}\\x+y=\frac{5+2\sqrt{6}}{6}\end{matrix}\right.\)

Bạn tự giải hệ tìm điểm rơi nếu thích, số xấu quá

2.

\(VT\ge\sqrt{\left(x+y+z\right)^2+\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2}\ge\sqrt{\left(x+y+z\right)^2+\frac{81}{\left(x+y+z\right)^2}}\)

Đặt \(x+y+z=t\Rightarrow0< t\le1\)

\(VT\ge\sqrt{t^2+\frac{81}{t^2}}=\sqrt{t^2+\frac{1}{t^2}+\frac{80}{t^2}}\ge\sqrt{2\sqrt{\frac{t^2}{t^2}}+\frac{80}{1^2}}=\sqrt{82}\)

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

3.

\(\frac{a^2}{b^5}+\frac{a^2}{b^5}+\frac{a^2}{b^5}+\frac{1}{a^3}+\frac{1}{a^3}\ge5\sqrt[5]{\frac{a^6}{b^{15}.a^6}}=\frac{5}{b^3}\)

Tương tự: \(\frac{3b^2}{c^5}+\frac{2}{b^3}\ge\frac{5}{a^3}\) ; \(\frac{3c^2}{d^5}+\frac{2}{c^3}\ge\frac{5}{d^3}\) ; \(\frac{3d^2}{a^5}+\frac{2}{d^2}\ge\frac{5}{a^3}\)

Cộng vế với vế và rút gọn ta được: \(3VT\ge3VP\)

Dấu "=" xảy ra khi và chỉ khi \(a=b=c=d=1\)

4.

ĐKXĐ: \(-2\le x\le2\)

\(y^2=\left(x+\sqrt{4-x^2}\right)^2\le2\left(x^2+4-x^2\right)=8\)

\(\Rightarrow y\le2\sqrt{2}\Rightarrow y_{max}=2\sqrt{2}\) khi \(x=\sqrt{2}\)

Mặt khác do \(\left\{{}\begin{matrix}x\ge-2\\\sqrt{4-x^2}\ge0\end{matrix}\right.\) \(\Rightarrow x+\sqrt{4-x^2}\ge-2\)

\(y_{min}=-2\) khi \(x=-2\)

Từ bđt Cauchy : \(a+b\ge2\sqrt{ab}\) ta suy ra được \(ab\le\frac{\left(a+b\right)^2}{4}\)

Áp dụng vào bài toán của bạn :

a/ \(y=\left(x+3\right)\left(5-x\right)\le\frac{\left(x+3+5-x\right)^2}{4}=...............\)

b/ Tương tự

c/ \(y=\left(x+3\right)\left(5-2x\right)=\frac{1}{2}.\left(2x+6\right)\left(5-2x\right)\le\frac{1}{2}.\frac{\left(2x+6+5-2x\right)^2}{4}=.............\)

d/ Tương tự

e/ \(y=\left(6x+3\right)\left(5-2x\right)=3\left(2x+1\right)\left(5-2x\right)\le3.\frac{\left(2x+1+5-2x\right)^2}{4}=.......\)

f/ Xét \(\frac{1}{y}=\frac{x^2+2}{x}=x+\frac{2}{x}\ge2\sqrt{x.\frac{2}{x}}=2\sqrt{2}\)

Suy ra \(y\le\frac{1}{2\sqrt{2}}\)

..........................

g/ Đặt \(t=x^2\) , \(t>0\) (Vì nếu t = 0 thì y = 0)

\(\frac{1}{y}=\frac{t^3+6t^2+12t+8}{t}=t^2+6t+\frac{8}{t}+12\)

\(=t^2+6t+\frac{8}{3t}+\frac{8}{3t}+\frac{8}{3t}+12\)

\(\ge5.\sqrt[5]{t^2.6t.\left(\frac{8}{3t}\right)^3}+12=.................\)

Từ đó đảo ngược y lại rồi đổi dấu \(\ge\) thành \(\le\)

Mấy cái dấu "=" anh tự xét.

Áp dụng BĐT AM-GM: \(VT=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge3\sqrt[3]{\frac{1}{abc}}=\frac{3}{\sqrt[3]{abc}}\ge\frac{3}{\frac{a+b+c}{3}}=\frac{9}{a+b+c}\)

a) Áp dụng: \(VT\ge\frac{\left(a+b+c\right)^2}{3}.\frac{9}{2\left(a+b+c\right)}=\frac{3}{2}\left(a+b+c\right)\)

b) \(P=3-\left(\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}\right)\le3-\frac{9}{x+y+z+3}=\frac{3}{4}\)

a/ \(\frac{x}{2}+\frac{18}{x}\ge2\sqrt{\frac{x}{2}.\frac{18}{x}}=...\)

b/ \(\frac{x}{2}+\frac{2}{x-1}=\frac{x-1}{2}+\frac{2}{x-1}+\frac{1}{2}\ge2\sqrt{\frac{x-1}{2}.\frac{2}{x-1}}+\frac{1}{2}=...\)

c/ \(\frac{3x}{2}+\frac{1}{x+1}=\frac{3\left(x+1\right)}{2}+\frac{1}{x+1}-\frac{3}{2}\ge2\sqrt{\frac{3\left(x+1\right)}{2}.\frac{1}{x+1}}-\frac{3}{2}=...\)

d/ \(\frac{x}{3}+\frac{5}{2x-1}=\frac{2x-1}{6}+\frac{5}{2x-1}+\frac{1}{6}\ge2\sqrt{\frac{2x-1}{6}.\frac{5}{2x-1}}+\frac{1}{6}=...\)

e/ \(\frac{x}{1-x}+\frac{5}{x}=\frac{x}{1-x}+\frac{5-5x+5x}{x}=\frac{x}{1-x}+\frac{5\left(1-x\right)}{x}+5\ge2\sqrt{\frac{x}{1-x}.\frac{5\left(1-x\right)}{x}}+5=...\)

f/ \(\frac{x^3+1}{x^2}=x+\frac{1}{x^2}=\frac{x}{2}+\frac{x}{2}+\frac{1}{x^2}\ge2\sqrt{\frac{x}{2}.\frac{x}{2}.\frac{1}{x^2}}=...\)

g/ \(\frac{x^2+4x+4}{x}=x+\frac{4}{x}+4\ge2\sqrt{x.\frac{4}{x}}+4=...\)

\(VT=\frac{\left(yz\right)^2}{x^2yz\left(y+z\right)}+\frac{\left(zx\right)^2}{xy^2z\left(z+x\right)}+\frac{\left(xy\right)^2}{xyz^2\left(x+y\right)}\)

\(VT=\frac{2\left(yz\right)^2}{xy+xz}+\frac{2\left(zx\right)^2}{xy+yz}+\frac{2\left(xy\right)^2}{xz+yz}\)

\(VT\ge\frac{2\left(xy+yz+zx\right)^2}{2\left(xy+yz+zx\right)}=xy+yz+zx\)

Dấu "=" xảy ra khi \(x=y=z=\frac{1}{\sqrt[3]{2}}\)

a/ \(y=\left(x+3\right)\left(5-x\right)\le\frac{1}{4}\left(x+3+5-x\right)^2=16\)

Dấu "=" xảy ra khi \(x+3=5-x\Leftrightarrow x=1\)

b/ \(y=x\left(6-x\right)\le\frac{1}{4}\left(x+6-x\right)^2=9\)

\("="\Leftrightarrow x=3\)

c/ \(y=\frac{1}{2}\left(2x+6\right)\left(5-2x\right)\le\frac{1}{8}\left(2x+6+5-2x\right)^2=\frac{121}{8}\)

\("="\Leftrightarrow x=-\frac{1}{4}\)

d/ \(y=\frac{1}{2}\left(2x+5\right)\left(10-2x\right)\le\frac{1}{8}\left(2x+5+10-2x\right)^2=\frac{225}{8}\)

\("="\Leftrightarrow x=\frac{5}{4}\)

e/ \(y=3\left(2x+1\right)\left(5-2x\right)\le\frac{3}{4}\left(2x+1+5-2x\right)^2=27\)

\("="\Leftrightarrow x=1\)

f/ \(\frac{x}{x^2+2}\le\frac{x}{2\sqrt{x^2.2}}=\frac{1}{2\sqrt{2}}\)

\("="\Leftrightarrow x=\sqrt{2}\)

g/ \(y=\frac{x^2}{\left(x^2+\frac{3}{2}+\frac{3}{2}\right)^3}\le\frac{x^2}{\left(3\sqrt[3]{\frac{9}{4}x^2}\right)^3}=\frac{4}{243}\)

\("="\Leftrightarrow x^2=\frac{3}{2}\Leftrightarrow x=\pm\sqrt{\frac{3}{2}}\)