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đặt a = 2x + y + z; b = 2y + z + x; c = 2z + x + y (a; b ; c > 0)
=> a + b + c = 4.(x+ y + z) => x + y + z = (a+ b+ c) / 4
=> x = a - (x+ y + z) = a - (a+ b + c) / 4
y = b - (x + y + z) = b - (a+b+c) / 4
z = c - (x+y + z) = c - (a+b+c)/ 4
Khi đó : \(VT=1-\frac{a+b+c}{4a}+1-\frac{a+b+c}{4b}+1-\frac{a+b+c}{4c}\)
\(VT=3-\left(\frac{a+b+c}{4a}+\frac{a+b+c}{4b}+\frac{a+b+c}{4c}\right)=3-\frac{1}{4}.\left(a+b+c\right).\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
\(VT=3-\frac{1}{4}.\left(1+\frac{a}{b}+\frac{a}{c}+\frac{b}{a}+1+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}+1\right)=3-\frac{1}{4}.\left(3+\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{a}{c}+\frac{c}{a}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)\right)\)
Với a, b > 0 ta có: a/b + b/ a > = 2
=> \(\frac{1}{4}.\left(3+\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{a}{c}+\frac{c}{a}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)\right)\ge\frac{1}{4}.\left(3+2+2+2\right)=\frac{9}{4}\)
=> \(VT\le3-\frac{9}{4}=\frac{3}{4}\)
Dấu = xảy ra khi a= b = c => x = y = z
Đặt \(a=2x+y+z;b=2y+z+x;c=2z+x+y\)
\( \implies\) \(a+b+c=\left(2x+y+z\right)+\left(2y+z+x\right)+\left(2z+x+y\right)\)
\( \implies\) \(a+b+c=4x+4y+4z\)
\( \implies\) \(x+y+z=\frac{a+b+c}{4}\)
+)Ta có : \(a=2x+y+z\)
\(\iff\) \(a=x+\left(x+y+z\right)\)
\(\iff\) \(a-\left(x+y+z\right)=x\)
\(\iff\) \(a-\frac{a+b+c}{4}=x\)
\(\iff\) \(x=\frac{3a-b-c}{4}\)
+)Ta có :\(b=2y+z+x\)
\(\iff\) \(b=y+\left(y+z+x\right)\)
\(\iff\)\(b-\left(y+z+x\right)=y\)
\(\iff\) \(b-\frac{a+b+c}{4}=y\)
\(\iff\)\(y=\frac{3b-c-a}{4}\)
+)Ta có :\(c=2z+x+y\)
\(\iff\) \(c=z+\left(z+x+y\right)\)
\(\iff\) \(c-\left(z+x+y\right)=z\)
\(\iff\) \(c-\frac{a+b+c}{4}=z\)
\(\iff\)\(z=\frac{3c-a-b}{4}\)
\( \implies\) \(\frac{x}{2x+y+z}+\frac{y}{2y+z+x}+\frac{z}{2z+x+y}\)
\(=\frac{3a-b-c}{4a}+\frac{3b-c-a}{4b}+\frac{3c-a-b}{4c}\)
\(=\frac{9}{4}-\left(\frac{b}{4a}+\frac{c}{4a}+\frac{c}{4b}+\frac{a}{4b}+\frac{a}{4c}+\frac{b}{4c}\right)\)
\(=\frac{9}{4}-\frac{1}{4}\left(\frac{b}{a}+\frac{c}{a}+\frac{c}{b}+\frac{a}{b}+\frac{a}{c}+\frac{b}{c}\right)\)
\(=\frac{9}{4}-\frac{1}{4}\left[\left(\frac{b}{a}+\frac{a}{b}\right)+\left(\frac{c}{a}+\frac{a}{c}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)\right]\)
Áp dụng bất đẳng thức ( BĐT Cosi ) : \(m+n\)\( \geq\)\(2\sqrt{mn}\) \(\left(m;n>0\right)\)ta được :
\(\frac{b}{a}+\frac{a}{b}\) \( \geq\) 2 \(\sqrt{\frac{b}{a}.\frac{a}{b}}\) = 2 \( \implies\) \(\frac{b}{a}+\frac{a}{b}\) \( \geq\) 2
\(\frac{c}{a}+\frac{a}{c}\) \( \geq\) 2 \(\sqrt{\frac{c}{a}.\frac{a}{c}}\) = 2 \( \implies\) \(\frac{c}{a}+\frac{a}{c}\) \( \geq\) 2
\(\frac{b}{c}+\frac{c}{b}\) \( \geq\) 2 \(\sqrt{\frac{b}{c}.\frac{c}{b}}\) = 2 \( \implies\) \(\frac{b}{c}+\frac{c}{b}\) \( \geq\) 2
\( \implies\) \(\left(\frac{b}{a}+\frac{a}{b}\right)+\left(\frac{c}{a}+\frac{a}{c}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)\) \( \geq\) 2 + 2 + 2
\( \implies\) \(\left(\frac{b}{a}+\frac{a}{b}\right)+\left(\frac{c}{a}+\frac{a}{c}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)\) \( \geq\) 6
\( \implies\) \(\frac{1}{4}\left[\left(\frac{b}{a}+\frac{a}{b}\right)+\left(\frac{c}{a}+\frac{a}{c}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)\right]\) \( \geq\) \(\frac{6}{4}\)
\( \implies\) \(\frac{1}{4}\left[\left(\frac{b}{a}+\frac{a}{b}\right)+\left(\frac{c}{a}+\frac{a}{c}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)\right]\) \( \geq\) \(\frac{3}{2}\)
\( \implies\) \(-\frac{1}{4}\left[\left(\frac{b}{a}+\frac{a}{b}\right)+\left(\frac{c}{a}+\frac{a}{c}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)\right]\) \(\leq\) \(-\frac{3}{2}\)
\( \implies\) \(\frac{9}{4}-\frac{1}{4}\left[\left(\frac{b}{a}+\frac{a}{b}\right)+\left(\frac{c}{a}+\frac{a}{c}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)\right]\) \(\leq\) \(\frac{9}{4}-\frac{3}{2}\)
\( \implies\) \(\frac{9}{4}-\frac{1}{4}\left[\left(\frac{b}{a}+\frac{a}{b}\right)+\left(\frac{c}{a}+\frac{a}{c}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)\right]\) \(\leq\) \(\frac{3}{4}\)
Cho các số nguyên dương x, y, z. Chứng minh rằng:
\(1< \frac{x}{x+y}+\frac{y}{y+z}+\frac{z}{z+x}< 2\)
Với x, y, z nguyên dương
Ta có: \(\frac{x}{x+y}>\frac{x}{x+y+z}\)
\(\frac{y}{y+z}>\frac{y}{x+y+z}\)
\(\frac{z}{z+x}>\frac{z}{x+y+z}\)
\(\Rightarrow\frac{x}{x+y}+\frac{y}{y+z}+\frac{z}{z+x}>\frac{x+y+z}{x+y+z}=1\)(1)
Mặt khác \(\frac{x}{x+y}< 1\Rightarrow\frac{x}{x+y}< \frac{x+z}{x+y+z}\)
\(\frac{y}{y+z}< \frac{y+x}{x+y+z}\)
\(\frac{z}{z+x}< \frac{z+y}{x+y+z}\)
\(\Rightarrow\frac{x}{x+y}+\frac{y}{y+z}+\frac{z}{z+x}< 2\)(2)
Từ (1) và (2) => dpcm
Có : x/x+y ; y/y+z ; z/z+x đều > 0
=> x/z+y + y/y+z + z/z+x > x/x+y+z + y/x+y+z + z/x+y+z = x+y+z/x+y+z = 1 (1)
Lại có : x,y,z > 0
=> 0 < x/x+y ; y/y+z ; z/z+x < 1
=> x/x+y + y/y+z + z/z+x < x+z/x+y+z + y+x/x+y+z + z+y/x+y+z = x+z+y+x+z+y/x+y+z = 2 (2)
Từ (1) và (2) => ĐPCM
Tk mk nha
* Với \(a=1\) ta thấy BĐT đúng.
* Ta xét khi \(a>1\)
Hàm nghi số \(y=\) \(y=\frac{1}{a^1}=\left(\frac{1}{a}\right)^1\) nghịch biến với \(\forall t\in R,\) khi \(a>1\).
Khi đó ta có
Ta có: \(\left(x-y\right)\left(\frac{1}{a^x}-\frac{1}{a^y}\right)\le0,\forall x,y\in R\Rightarrow\frac{x}{a^x}+\frac{y}{a^y}\le\frac{x}{a^y}+\frac{y}{a^x}\) (1)
Chứng minh tương tự \(\frac{y}{a^y}+\frac{z}{a^z}\le\frac{z}{a^y}+\frac{y}{a^z}\) (2) \(\frac{z}{a^z}+\frac{x}{a^x}\le\frac{x}{a^z}+\frac{z}{a^x}\) (3)
Cộng vế với vế (1), (2) và (3) ta được \(2\left(\frac{x}{a^x}+\frac{y}{a^y}+\frac{z}{a^z}\right)\le\frac{y+z}{a^x}+\frac{z+x}{a^y}+\frac{x+y}{a^z}\) (4)
Cộng 2 vế của (4) với biểu thức \(\frac{x}{a^x}+\frac{y}{a^y}+\frac{z}{a^z}\) ta được
\(3\left(\frac{x}{a^x}+\frac{y}{a^y}+\frac{z}{a^z}\right)\le\frac{x+y+z}{a^x}+\frac{x+y+z}{a^y}+\frac{x+y+z}{a^z}=\left(x+y+z\right)\left(\frac{1}{a^x}+\frac{1}{a^y}+\frac{1}{a^z}\right)\)
Áp dụng bđt Cauchy - Schwarz dạng Engel:
\(\frac{1}{2x}+\frac{1}{2y}\ge\frac{4}{2\left(x+y\right)}=\frac{2}{x+y}\)
Dấu "=" xảy ra khi x = y > 0
thanks