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\(\Leftrightarrow P\left(x\sqrt{y}+y\sqrt{z}+z\sqrt{x}\right)\ge\left(x+y+z\right)^2\left(1\right)\)
Áp dụng Bu-nhi :
\(\left(x\sqrt{y}+y\sqrt{z}+z\sqrt{x}\right)^2\le\left(xy+yz+xz\right)\left(x+y+z\right)\)
\(\Leftrightarrow x\sqrt{y}+y\sqrt{z}+z\sqrt{x}\le24\)
\(\Leftrightarrow P\left(x\sqrt{y}+y\sqrt{z}+z\sqrt{x}\right)\le24P\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\)\(\left(x+y+z\right)^2\le24P\)
\(\Rightarrow12^2\le24P\)
\(\Rightarrow P\ge6\)
ĐẾN ĐÂY BẠN TỰ GIẢI DẤU \(=\) XẢY RA LÚC NÀO NHÉ
Áp dụng Bu-nhi :
\(12^2<\left(x+y+z\right)^2=\left(\frac{\sqrt{x}}{\sqrt{\sqrt{y}}}.\sqrt{x}.\sqrt{\sqrt{y}}+\frac{\sqrt{y}}{\sqrt{\sqrt{z}}}.\sqrt{y}.\sqrt{\sqrt{z}}+\frac{\sqrt{z}}{\sqrt{\sqrt{x}}}.\sqrt{z}.\sqrt{\sqrt{x}}\right)^2\)
\(\le\left(\frac{x}{\sqrt{y}}+\frac{y}{\sqrt{z}}+\frac{z}{\sqrt{x}}\right)\left(x\sqrt{y}+y\sqrt{z}+z\sqrt{x}\right)\)
Bài 1:
Đặt \(a^2=x;b^2=y;c^2=z\)
Ta có:\(\sqrt{\frac{x}{x+y}}+\sqrt{\frac{y}{y+z}}+\sqrt{\frac{z}{z+x}}\le\frac{3}{\sqrt{2}}\)
Áp dụng BĐT cô si ta có:
\(\sqrt{\frac{x}{x+y}}=\frac{1}{\sqrt{2}}\sqrt{\frac{4x\left(x+y+z\right)}{3\left(x+y\right)\left(x+z\right)}\frac{3\left(x+z\right)}{2\left(x+y+z\right)}}\)
\(\le\frac{1}{2\sqrt{2}}\left[\frac{4x\left(x+y+z\right)}{3\left(x+y\right)\left(x+z\right)}+\frac{3\left(x+z\right)}{2\left(x+y+z\right)}\right]\)
Tương tự với \(\sqrt{\frac{y}{y+z}}\)và \(\sqrt{\frac{z}{z+x}}\)
Cộng lại ta được:
\(\frac{\sqrt{2}}{3}\left[\frac{x\left(x+y+z\right)}{\left(x+y\right)\left(x+z\right)}+\frac{y\left(x+y+z\right)}{\left(y+z\right)\left(y+x\right)}+\frac{z\left(x+y+z\right)}{\left(z+x\right)\left(z+y\right)}\right]+\frac{3}{2\sqrt{2}}\le\frac{3}{2\sqrt{2}}\)
Sau đó bình phương hai vế rồi
\(\Rightarrow\left(x+y\right)\left(y+z\right)\left(z+x\right)\ge8xyz\)đẳng thức đúng
Vậy...
Bài 2:
Trước hết ta chứng minh bất đẳng thức sau:
\(\frac{a}{4a+4b+c}+\frac{b}{4b+4c+a}+\frac{c}{4c+4a+b}\le\frac{1}{3}\)
Nhân cả hai vế bđt với 4(a+b+c)4(a+b+c) rồi thu gọn ta được bđt sau:
\(\frac{4a\left(a+b+c\right)}{4a+4b+c}+\frac{4b\left(a+b+c\right)}{4b+4c+a}+\frac{4c\left(a+b+c\right)}{4c+4a+b}\)\(\le\frac{4}{3}\left(a+b+c\right)\)
\(\left[\frac{4a\left(a+b+c\right)}{4a+4b+}-a\right]+\left[\frac{4b\left(a+b+c\right)}{4b+4c+a}-b\right]+\left[\frac{4c\left(a+b+c\right)}{4c+4a+b}-c\right]\le\frac{a+b+c}{3}\)
\(\frac{ca}{4a+4b+c}+\frac{ab}{4b+4c+a}+\frac{bc}{4c+4a+b}\le\frac{a+b+c}{9}\)
Áp dụng bđt cauchy-Schwarz ta có \(\frac{ca}{4a+4b+c}=\frac{ca}{\left(2b+c\right)+2\left(2a+b\right)}\)\(\le\frac{ca}{9}\left(\frac{1}{2b+c}+\frac{2}{2a+b}\right)\)
Từ đó ta có:
\(\text{∑}\frac{ca}{4a+4b+c}\le\frac{1}{9}\text{∑}\left(\frac{ca}{2b+c}+\frac{2ca}{2a+b}\right)\)\(=\frac{1}{9}\left(\text{ ∑}\frac{ca}{2b+c}+\text{ ∑}\frac{2ca}{2a+b}\right)\)\(=\frac{1}{9}\left(\text{ ∑}\frac{ca}{2b+c}+\text{ ∑}\frac{2ab}{2b+c}\right)=\frac{a+b+c}{9}\)
Đặt VT=A rồi áp dụng bđt cauchy-Schwarz cho VT ta có
\(T^2\le3\left(\frac{a}{4a+4b+c}+\frac{b}{4b+4c+a}+\frac{c}{4c+4a+b}\right)\)\(\le3\cdot\frac{1}{3}=1\Leftrightarrow T\le1\)
Dấu = xảy ra khi a=b=c
c bạn tự làm nhé mình mệt rồi :D
Ta có: \(\sqrt{\frac{b+c}{a}}\le\frac{1+\frac{b+c}{a}}{2}=\frac{a+b+c}{2a}\)
\(\Rightarrow\sqrt{\frac{a}{b+c}}\ge\frac{2a}{a+b+c}\)
Tương tự \(\sqrt{\frac{b}{c+a}}\ge\frac{2b}{a+b+c};\sqrt{\frac{c}{a+b}}\ge\frac{2c}{a+b+c}\)
\(\Rightarrow\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}+\sqrt{\frac{c}{a+b}}\ge2\)
Dấu "=" xảy ra khi a=b=c=0 (trái gt)
\(\Rightarrow\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}+\sqrt{\frac{c}{a+b}}>2\)(đpcm)
3.Áp dụng BĐT \(\frac{1}{x+y+z}\le\frac{1}{9}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)ta có
\(\frac{ab}{a+3b+2c}=ab.\frac{1}{\left(a+c\right)+2b+\left(b+c\right)}\le\frac{1}{9}ab.\left(\frac{1}{a+c}+\frac{1}{2b}+\frac{1}{b+c}\right)\)
TT \(\frac{bc}{b+3c+2a}\le\frac{bc}{9}.\left(\frac{1}{b+a}+\frac{1}{2c}+\frac{1}{c+a}\right)\)
\(\frac{ca}{c+3a+2b}\le\frac{ac}{9}.\left(\frac{1}{a+b}+\frac{1}{2a}+\frac{1}{b+c}\right)\)
=> \(VT\le\frac{1}{18}\left(a+b+c\right)+\Sigma.\frac{1}{9}.\left(\frac{bc}{a+c}+\frac{ba}{a+c}\right)=\frac{1}{18}\left(a+b+c\right)+\frac{1}{9}\left(a+b+c\right)=\frac{1}{6}\left(a+b+c\right)\)
Dấu bằng xảy ra khi a=b=c
cảm ơn bạn nhiều, bạn có thể giúp mình hai câu kia nữa được không
c/ \(C'=\frac{1}{\frac{1}{3-2\sqrt{x}}}.\frac{1}{\frac{1}{\sqrt{3-2\sqrt{x}}}+1}=\frac{\sqrt{\left(3-2\sqrt{x}\right)^3}}{1+\sqrt{\left(3-2\sqrt{x}\right)}}\)
Đặt \(\sqrt{\left(3-2\sqrt{x}\right)}=a\)
\(\Rightarrow C'=\frac{a^3}{a+1}=a^2-a+1-\frac{1}{a+1}\)
Đế C' nguyên thì a + 1 là ước của 1
\(\Rightarrow a=0\)
\(\Rightarrow\sqrt{\left(3-2\sqrt{x}\right)}=0\)
\(\Rightarrow x=\frac{9}{4}\left(l\right)\)
Vậy không có x.
Không biết có nhầm chỗ nào không nữa. Lam biếng kiểm tra lại quá. You kiểm tra lại hộ nhé. Thanks
a/ \(C=\left(\frac{2\sqrt{x}}{2x-5\sqrt{x}+3}-\frac{5}{2\sqrt{x}-3}\right):\left(3+\frac{2}{1-\sqrt{x}}\right)\)
\(=\left(\frac{2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(2\sqrt{x}-3\right)}-\frac{5}{2\sqrt{x}-3}\right):\left(\frac{3\sqrt{x}-5}{\sqrt{x}-1}\right)\)
\(=\left(\frac{2\sqrt{x}-5\sqrt{x}+5}{\left(\sqrt{x}-1\right)\left(2\sqrt{x}-3\right)}\right):\left(\frac{3\sqrt{x}-5}{\sqrt{x}-1}\right)\)
\(=\frac{5-3\sqrt{x}}{\left(\sqrt{x}-1\right)\left(2\sqrt{x}-3\right)}.\frac{\sqrt{x}-1}{3\sqrt{x}-5}\)
\(=\frac{1}{3-2\sqrt{x}}\)
Câu b, c tự làm nhé
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