Ta có :

\(A=\sqrt{\frac{x^3}{x^3+8y^3}}\)

\(\Rightarrow A=\sqrt{\frac{1}{1+\left(\frac{2y}{x}\right)^3}}\)

\(\Rightarrow A=\sqrt{\frac{1}{\left(1+\frac{2y}{x}\right)\left(1-\frac{2y}{x}+\frac{4y^2}{x^2}\right)}}\)

\(\Rightarrow A\ge\frac{1}{\frac{\left(1+\frac{2y}{x}\right)+\left(1-\frac{2y}{x}+\frac{4y^2}{x^2}\right)}{2}}\)

\(\Rightarrow A\ge\frac{2}{2+\frac{4y^2}{x^2}}=\frac{1}{1+2\left(\frac{y}{x}\right)^2}\)

VÀ

\(B=\sqrt{\frac{4y^3}{y^3+\left(x+y\right)^3}}\)

\(\Rightarrow B=\sqrt{\frac{4}{1+\left(\frac{x}{y}+1\right)^3}}\)

\(\Rightarrow B=\frac{2}{\sqrt{\left[1+\left(1+\frac{x}{y}\right)\right]\left[1-\left(1+\frac{x}{y}\right)+\left(1+\frac{x}{y}\right)^2\right]}}\)

\(\Rightarrow B\ge\frac{2}{\frac{\left[1+\left(1+\frac{x}{y}\right)\right]+\left[1-\left(1+\frac{x}{y}\right)+\left(1+\frac{x}{y}\right)^2\right]}{2}}\)

\(\Rightarrow B\ge\frac{4}{2+\left(1+\frac{x}{y}\right)^2}\)

Suy ra :

\(P=A+B\ge\frac{1}{1+2\left(\frac{y}{x}\right)^2}+\frac{4}{2+\left(1+\frac{x}{y}\right)^2}\)

\(\Rightarrow P\ge\frac{x^2}{x^2+2y^2}+\frac{4y^2}{2y^2+\left(x+y\right)^2}\)

\(\Rightarrow P\ge\frac{x^2}{x^2+2y^2}+\frac{4y^2}{2y^2+2\left(x^2+y^2\right)}=\frac{x^2}{x^2+2y^2}+\frac{4y^2}{2x^2+4y^2}=\frac{x^2}{x^2+2y^2}+\frac{2y^2}{x^2+2y^2}=1\)

"=" khi \(x=y\)

lô bn xàm lồn

bn trẩu , m phải ARMY hơm 

nếu phải thì nhục quá trời,tự nhiên fan BTS lại chưa con phò như mài ,u hú hú

bớt sàm lại đuy,ko thì đừng làm AMI nx,ư~~

Bài 1: \(T=\sqrt{\frac{x^3}{x^3+8y^3}}+\sqrt{\frac{4y^3}{y^3+\left(x+y\right)^3}}\)

\(=\frac{x^2}{\sqrt{x\left(x^3+8y^3\right)}}+\frac{2y^2}{\sqrt{y\left[y^3+\left(x+y\right)^3\right]}}\)

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

\(\ge\frac{2x^2}{2x^2+4y^2}+\frac{4y^2}{2y^2+\left(x+y\right)^2}\ge\frac{2x^2}{2x^2+4y^2}+\frac{4y^2}{2x^2+4y^2}=1\)

\(\Rightarrow T\ge1\)

Bài 2:

[Toán 10] Bất đẳng thức | Page 5 | HOCMAI Forum - Cộng đồng học sinh Việt Nam

hình như có dấu + giữa 2 phân số

Đúng rồi Thắng , bài này đúng ra phải là \(A=\sqrt{\frac{x^3}{x^3+8y^3}}+\sqrt{\frac{4y^3}{y^3+\left(x+y\right)^3}}\)

\(A=\frac{x^2}{\sqrt{x^4+8xy^3}}+\frac{2y^2}{\sqrt{y\left(y^3+\left(x+y\right)^3\right)}}\)

Áp dụng BĐT Cauchy, ta có: 

\(x^4+8xy^3=x^4+8.xy.y^2\le x^4+4\left(x^2y^2+y^4\right)=\left(x^2+2y^2\right)^2\)

\(\Rightarrow\frac{x^2}{\sqrt{x^3+8xy^3}}\ge\frac{x^2}{x^2+2y^2}\)

\(\sqrt{y\left(y^3+\left(x+y\right)^3\right)}=\sqrt{\left(xy+2y^2\right)\left(x^2+y^2+xy\right)}\le\frac{x^2+3y^2+2xy}{2}=\frac{2y^2+\left(x+y\right)^2}{2}\)

\(\le\frac{2y^2+2\left(x^2+y^2\right)}{2}=x^2+2y^2\)

\(\Rightarrow A\ge\frac{x^2}{x^2+2y^2}+\frac{2y^2}{x^2+2y^2}=1\)

Vậy minA = 1 tại x = y > 0

post từng câu một thôi bn nhìn mệt quá

Lời giải:

Đặt $\frac{y}{x}=a(a>0)$ thì:

\(P=\sqrt{\frac{1}{1+(\frac{2y}{x})^3}}+\sqrt{\frac{4}{1+(1+\frac{x}{y})^3}}=\sqrt{\frac{1}{1+8a^3}}+\sqrt{\frac{4}{1+(1+\frac{1}{a})^3}}\)

Áp dụng BĐT AM-GM dạng $xy\leq \left(\frac{x+y}{2}\right)^2$ ta có:

\(1+8a^3=1+(2a)^3=(1+2a)(1-2a+4a^2)\leq \left(\frac{1+2a+1-2a+4a^2}{2}\right)^2=(2a^2+1)^2\)

\(\Rightarrow \sqrt{\frac{1}{8a^3+1}}\geq \frac{1}{2a^2+1}(1)\)

\(1+(1+\frac{1}{a})^3=(2+\frac{1}{a})[1-(1+\frac{1}{a})+(1+\frac{1}{a})^2]\leq (\frac{3a^2+2a+1}{2a^2})^2\)

\(\Rightarrow \sqrt{\frac{4}{1+(1+\frac{1}{a})^3}}\geq \frac{4a^2}{3a^2+2a+1}\)

Mà: \(\frac{4a^2}{3a^2+2a+1}\geq \frac{4a^2}{3a^2+a^2+1+1}=\frac{2a^2}{2a^2+1}\) nên \(\sqrt{\frac{4}{1+(1+\frac{1}{a})^3}}\geq \frac{2a^2}{2a^2+1}(2)\)

Từ $(1);(2)\Rightarrow P\geq \frac{1}{2a^2+1}+\frac{2a^2}{2a^2+1}=1$

Vậy $P_{\min}=1$ khi $a=1\Leftrightarrow x=y$

sorry, mìh mới học lớp seven thôi

Bài 1: diendantoanhoc.net

Đặt \(a=\frac{1}{x};b=\frac{1}{y};c=\frac{1}{z}\) BĐT cần chứng minh trở thành

\(\frac{x}{\sqrt{3zx+2yz}}+\frac{x}{\sqrt{3xy+2xz}}+\frac{x}{\sqrt{3yz+2xy}}\ge\frac{3}{\sqrt{5}}\)

\(\Leftrightarrow\frac{x}{\sqrt{5z}\cdot\sqrt{3x+2y}}+\frac{y}{\sqrt{5x}\cdot\sqrt{3y+2z}}+\frac{z}{\sqrt{5y}\cdot\sqrt{3z+2x}}\ge\frac{3}{5}\)

Theo BĐT AM-GM và Cauchy-Schwarz ta có:

\( {\displaystyle \displaystyle \sum }\)\(_{cyc}\frac{x}{\sqrt{5z}\cdot\sqrt{3x+2y}}\ge2\)\( {\displaystyle \displaystyle \sum }\)\(\frac{x}{3x+2y+5z}\ge\frac{2\left(x+y+z\right)^2}{x\left(3x+2y+5z\right)+y\left(5x+3y+2z\right)+z\left(2x+5y+3z\right)}\)

\(=\frac{2\left(x+y+z\right)^2}{3\left(x^2+y^2+z^2\right)+7\left(xy+yz+zx\right)}\)

\(=\frac{2\left(x+y+z\right)^2}{3\left(x^2+y^2+z^2\right)+\frac{1}{3}\left(xy+yz+zx\right)+\frac{20}{3}\left(xy+yz+zx\right)}\)

\(\ge\frac{2\left(x+y+z\right)^2}{3\left(x^2+y^2+z^2\right)+\frac{1}{3}\left(x^2+y^2+z^2\right)+\frac{20}{3}\left(xy+yz+zx\right)}\)

\(=\frac{2\left(x^2+y^2+z^2\right)}{5\left[x^2+y^2+z^2+2\left(xy+yz+zx\right)\right]}=\frac{3}{5}\)

Bổ sung bài 1:

BĐT được chứng minh

Đẳng thức xảy ra <=> a=b=c

\(T=\sqrt{\dfrac{x^3}{x^3+8y^3}}+\sqrt{\dfrac{4y^3}{y^3+\left(x+y\right)^3}}\)

\(=\dfrac{x^2}{\sqrt{x\left(x^3+8y^3\right)}}+\dfrac{2y^2}{\sqrt{y\left(y^3+\left(x+y\right)^3\right)}}\)

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

\(\ge\dfrac{2x^2}{2x^2+4y^2}+\dfrac{4y^2}{2y^2+\left(x+y\right)^2}\)\(\ge\dfrac{2x^2}{2x^2+4y^2}+\dfrac{4y^2}{4y^2+2x^2}\)

\(\ge\dfrac{2x^2+4y^2}{2x^2+4y^2}=1\)