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Áp dụng bđt \(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\left(x;y>0\right)\) (tự c/m ha)
\(\frac{7}{a}+\frac{5}{b}+\frac{4}{c}=\left(\frac{4}{a}+\frac{4}{b}\right)+\left(\frac{1}{b}+\frac{1}{c}\right)+\left(\frac{3}{a}+\frac{3}{c}\right)\)
\(=4\left(\frac{1}{a}+\frac{1}{b}\right)+\left(\frac{1}{b}+\frac{1}{c}\right)+3\left(\frac{1}{a}+\frac{1}{c}\right)\)
\(\ge4.\frac{4}{a+b}+\frac{4}{b+c}+3.\frac{4}{a+c}=4\left(\frac{4}{a+b}+\frac{1}{b+c}+\frac{3}{c+a}\right)\)
Dấu "=" <=> a = b = c
tự tìm đkxđ
\(\Leftrightarrow\left(4x^3-8x^2+4x\right)+\left(-17x^2+39x-22\right)+\left(x+\sqrt{3x-2}-\sqrt{3x-2}\right)=0\)
\(\Leftrightarrow4x.\left(x-1\right)^2+\left(x-1\right)\left(-17x+22\right)+\sqrt{3x-2}\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(4x^2-4x-17x+22+\sqrt{3x-2}\right)=0\)
\(\Leftrightarrow x=1\) tự chứng minh vế kia >=0 đi :D
a + b + c = 3
a/( 1 + b^2 ) + b/( 1 + c^2 ) + c/( 1 + a^2 ) ≥ 3/2
Ta có
a/( 1 + b^2 ) = a - ab^2/( 1 + b^2 ) ≥ a - ab^2/2b = a - ab/2
Tương tự ta có
b/( 1 + c^2 ) ≥ b - bc/2
c/( 1 + a^2 ) ≥ c - ac/2
Cộng vào ta có
a/( 1 + b^2 ) + b/( 1 + c^2 ) + c/( 1 + a^2 ) ≥ a + b + c - ( ab + bc + ac )/2 = 3 - ( ab + bc + ac )/2
Xét ab + bc + ac
Ta có
a^2 + b^2 ≥ 2ab
b^2 + c^2 ≥ 2bc
c^2 + a^2 ≥ 2ac
=> a^2 + b^2 + c^2 ≥ ab + bc + ac
<=> a^2 + b^2 + c^2 + 2ac + 2bc + 2ab ≥ 3( ab + ac + bc )
<=> ( a + b + c )^2 ≥ 3( ab + ac + bc )
<=> ab + ac + bc ≤ 9:3 = 3
=> 3 - ( ab + bc + ac )/2 ≥ 3 - 3/2 = 3/2
=> a/( 1 + b^2 ) + b/( 1 + c^2 ) + c/( 1 + a^2 ) ≥
Theo dãy tỉ số bằng nhau , có :
\(\frac{a}{b+c}=\frac{b}{c+a}=\frac{c}{a+b}=\frac{a+b+c}{b+c+c+a+a+b}=\frac{1}{2}\)
\(\Rightarrow\frac{b+c}{a}=\frac{c+a}{b}=\frac{a+b}{c}=2\)
\(\Rightarrow\left(\frac{b+c}{a}\right)^3=\left(\frac{c+a}{b}\right)^3=\left(\frac{a+b}{c}\right)^3=8\)
\(\Rightarrow\frac{\left(b+c\right)^3}{a^3}+\frac{\left(c+a\right)^3}{b^3}+\frac{\left(a+b\right)^3}{c^3}=8.3=24\)
Bài này có trong đề thi HSG 9 của huyện hay tỉnh nào đấy :)) được cái thầy t bắt cày đi cày lại cả chục cái đề thi nên bài này t nhớ lắm :))
Với x là số dương, áp dụng bđt Cô-si
\(\sqrt{x^3+1}=\sqrt{\left(x+1\right)\left(x^2-x+1\right)}\le\frac{x+1+x^2-x+1}{2}=\frac{x^2+2}{2}\)
\(\Rightarrow\sqrt{\frac{1}{x^3}}\ge\frac{2}{x^2+2}\) (*)
Dấu (=) xảy ra khi x = 2
Áp dụng bđt (*)
\(\sqrt{\frac{a^3}{a^3+\left(b+c\right)^3}}=\sqrt{\frac{1}{1+\left(\frac{b+c}{a}\right)^3}}\ge\frac{2}{\left(\frac{b+c}{a}\right)^2+2}=\frac{2a^2}{\left(b+c\right)^2+2a^2}\)
\(\Rightarrow\sqrt{\frac{a^3}{a^3+\left(b+c\right)^3}}\ge\frac{2a^2}{2\left(b^2+c^2\right)+2a^2}=\frac{a^2}{a^2+b^2+c^2}\left(1\right)\)
CMTT :
\(\sqrt{\frac{b^3}{b^3+\left(a+c\right)^3}}\ge\frac{b^2}{a^2+b^2+c^2}\) (2)
\(\sqrt{\frac{c^3}{c^3+\left(a+b\right)^3}}\ge\frac{c^2}{a^2+b^2+c^2}\) (3)
Cộng vế với vế của (1) ; (2) ; (3) ; ta được ĐPCM
\(\sqrt{\frac{1}{x^3}}\ge\frac{2}{x^2+2}\Rightarrow\sqrt{\frac{1}{1+\left(\frac{b+c}{a}\right)^3}}\ge\frac{2}{\left(\frac{b+c}{a}\right)^2+2}\)
Có nhầm chỗ nào ko vậy bạn chứ ở dưới mẫu có cộng 1 nữa mà
Có: \(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+3\ge4\left(\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}\right)\)
<=> \(\left(\frac{a}{b}+1\right)+\left(\frac{b}{c}+1\right)+\left(\frac{c}{a}+1\right)\ge\frac{4a}{a+b}+\frac{4b}{b+c}+\frac{4c}{c+a}\)
<=> \(\frac{a+b}{b}+\frac{b+c}{c}+\frac{c+a}{a}\ge\frac{4a}{a+b}+\frac{4b}{b+c}+\frac{4c}{c+a}\)
<=> \(\frac{a+b}{b}+\frac{b+c}{c}+\frac{c+a}{a}+\frac{4b}{a+b}+\frac{4c}{b+c}+\frac{4a}{c+a}\)
\(\ge\frac{4a}{a+b}+\frac{4b}{b+c}+\frac{4c}{c+a}+\frac{4b}{a+b}+\frac{4c}{b+c}+\frac{4a}{c+a}\)
<=> \(\left(\frac{a+b}{b}+\frac{4b}{a+b}\right)+\left(\frac{b+c}{c}+\frac{4c}{b+c}\right)+\left(\frac{c+a}{a}+\frac{4a}{c+a}\right)\)
\(\ge\left(\frac{4a}{a+b}+\frac{4b}{a+b}\right)+\left(\frac{4b}{b+c}+\frac{4c}{b+c}\right)+\left(\frac{4c}{c+a}+\frac{4a}{c+a}\right)\)
<=> \(\left(\frac{a+b}{b}+\frac{4b}{a+b}\right)+\left(\frac{b+c}{c}+\frac{4c}{b+c}\right)+\left(\frac{c+a}{a}+\frac{4a}{c+a}\right)\ge4+4+4\)
<=> \(\left(\frac{a+b}{b}+\frac{4b}{a+b}\right)+\left(\frac{b+c}{c}+\frac{4c}{b+c}\right)+\left(\frac{c+a}{a}+\frac{4a}{c+a}\right)\ge12\)(1)
Áp dụng Cô-si: (1) đúng.
Vậy Bất đẳng thức ban đầu đúng.
"=" <=> a = b = c.
\(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+3\ge4\left(\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{a+c}\right)\)
\(\Leftrightarrow\left(\frac{a}{b}+1\right)+\left(\frac{b}{c}+1\right)+\left(\frac{c}{a}+1\right)\ge4\left(\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}\right)\)
\(\Leftrightarrow\frac{a+b}{b}-\frac{4a}{a+b}+\frac{b+c}{c}-\frac{4b}{b+c}+\frac{c+a}{a}-\frac{4c}{c+a}\ge0\)
\(\Leftrightarrow\frac{\left(a-b\right)^2}{b\left(a+b\right)}+\frac{\left(b-c\right)^2}{c\left(b+c\right)}+\frac{\left(c-a\right)^2}{a\left(a+c\right)}\ge0\)
Luôn đúng vì a,b,c là các số dương
Dấu "=" xảy ra <=> a=b=c
Ta có : \(\left(x+y\right)^2\ge4xy\Rightarrow\frac{x+y}{xy}\ge\frac{4}{x+y}\)
\(\Rightarrow\frac{4}{x+y}\le\frac{1}{x}+\frac{1}{y}\)
\(\Rightarrow\frac{1}{x+y}\le\frac{1}{4}\left(\frac{1}{x}+\frac{1}{y}\right)\)
Áp dụng ta có :
\(\frac{a}{b+c}=a.\frac{1}{b+c}\le a.\frac{1}{4}\left(\frac{1}{b}+\frac{1}{c}\right)=\frac{1}{4}\left(\frac{a}{b}+\frac{a}{c}\right)\)
Tương tự :
\(\frac{b}{c+a}\le\frac{1}{4}\left(\frac{b}{c}+\frac{b}{a}\right)\)
\(\frac{c}{a+b}\le\frac{1}{4}\left(\frac{c}{a}+\frac{c}{b}\right)\)
\(\Rightarrow\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\le\frac{1}{4}\left(\frac{a}{b}+\frac{a}{c}\right)+\frac{1}{4}\left(\frac{b}{c}+\frac{b}{a}\right)+\frac{1}{4}\left(\frac{c}{a}+\frac{c}{b}\right)\)
\(\Rightarrow\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\le\frac{1}{4}\left(\frac{a}{b}+\frac{a}{c}+\frac{b}{c}+\frac{b}{a}+\frac{c}{a}+\frac{c}{b}\right)\)
\(\Rightarrow4\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)\le\frac{a}{b}+\frac{a}{c}+\frac{b}{c}+\frac{b}{a}+\frac{c}{a}+\frac{c}{b}\)
\(\Rightarrow4\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)\le\frac{a+c}{b}+\frac{a+b}{c}+\frac{b+c}{a}\)
\(\Rightarrow4\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)\le\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b}\)
Dấu = xảy ra khi a=b=c
Áp dụng BĐT cô si ta có :
\(\frac{b+c}{a}\ge4.\frac{a}{b+c}\)
\(\frac{c+a}{b}\ge\frac{4b}{c+a}\)
\(\frac{a+b}{c}\ge\frac{ac}{a+b}\)
\(\Rightarrow\frac{b+c}{a}+\frac{c+a}{b}+\frac{a+b}{c}\ge4.\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)\)
Dấu " = " xảy ra khi a= b = c
Đặt \(\left(\frac{a}{b+c};\frac{b}{c+a};\frac{c}{a+b}\right)\rightarrow\left(x;y;z\right)\) Khi đó ta có:
\(\left(x+y+z\right)^2+14xyz\ge4\)
Theo BĐT Nesbit \(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge\frac{3}{2}\Rightarrow x+y+z\ge\frac{3}{2}\)
\(VT=\left(x+y+z\right)^2+14xyz=x^2+y^2+z^2+2\left(xy+yz+xz\right)+14xyz\)
\(=x^2+y^2+z^2+6xyz+2\left(xy+yz+xz\right)+8xyz\)
\(\ge x^2+y^2+z^2+\frac{9xyz}{x+y+z}+2\left(xy+yz+xz\right)+8xyz\)
\(\ge4\left(xy+yz+xz\right)+8xyz=4\)
Ta có: \(\frac{a}{b}+1=\frac{a+b}{b}\)
*Cần c/m \(\frac{a+b}{b}\ge\frac{4a}{a+b},\forall a>0;b>0\) (*)
Thật vậy: (*)\(\Leftrightarrow\left(a+b\right)^2\ge4ab\Leftrightarrow...\Leftrightarrow\left(a-b\right)^2\ge0\) (luôn đúng với mọi a,b là số dương)
Tương tự ta cũng có: \(\frac{b}{c}+1=\frac{b+c}{c}\ge\frac{4b}{b+c}\); \(\frac{c}{a}+1=\frac{c+a}{a}\ge\frac{4c}{c+a}\)
Cộng theo vế ta được:
\(\left(\frac{a}{b}+1\right)+\left(\frac{b}{c}+1\right)+\left(\frac{c}{a}+1\right)\ge\frac{4a}{a+b}+\frac{4b}{b+c}+\frac{4c}{c+a}\)
\(\Leftrightarrow\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+3\ge4\left(\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}\right)\)