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Áp dụng BĐT Cô - si : x2 + y2 ≥ 2xy
=> \(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}\) ≥ \(2.\dfrac{a}{c}\) ( 1)
\(\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\) ≥ \(2.\dfrac{b}{a}\) ( 2)
\(\dfrac{a^2}{b^2}+\dfrac{c^2}{a^2}\) ≥ \(2.\dfrac{c}{b}\) ( 3)
Cộng từng vế của ( 1 , 3 , 3) , ta có :
\(2\left(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\right)\) ≥ \(2.\left(\dfrac{b}{a}+\dfrac{c}{b}+\dfrac{a}{c}\right)\)
=> ĐPCM
\(VT=\left(\dfrac{a}{b+c}+1\right)+\left(\dfrac{b}{c+a}+1\right)+\left(\dfrac{c}{a+b}+1\right)-3\)
\(=\dfrac{1}{2}\left(a+b\right)\left(b+c\right)\left(a+c\right)\left(\dfrac{1}{b+c}+\dfrac{1}{a+c}+\dfrac{1}{a+b}\right)-3>=\dfrac{9}{2}-3=\dfrac{3}{2}\)
Áp dụng BĐT Cô si dạng phân số ta có :
\(\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}\ge\dfrac{\left(a+b+c\right)^2}{2\left(a+b+c\right)}=\dfrac{a+b+c}{2}\)
=> ĐPCM .
b) Vì a,b,c > 0 .
Áp dụng BĐT Cô si ta có :
\(\dfrac{a^2}{b}+b\ge2a\) (1)
Tương tự ta có : \(\dfrac{b^2}{c}+c\ge2b\) (2)
\(\dfrac{c^2}{a}+a\ge2c\) (3)
Cộng từng vế => ĐPCM .
Áp dụng bđt Cauchy Schwarz dạng Engel ta có:
\(\left(a+b+c\right)\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{a+c}\right)\ge\left(a+b+c\right).\dfrac{\left(1+1+1\right)^2}{2\left(a+b+c\right)}\)
\(\ge\dfrac{9}{2}\left(đpcm\right)\)
Lời giải:
Mặc định đk $a,b,c\neq 0$
Áp dụng BĐT Cô-si cho các số dương ta có:
\(\frac{a^2}{b^2}+\frac{b^2}{c^2}\geq 2\sqrt{\frac{a^2}{b^2}.\frac{b^2}{c^2}}=2|\frac{a}{c}|\geq \frac{2a}{c}\)
\(\frac{a^2}{b^2}+\frac{c^2}{a^2}\geq 2\sqrt{\frac{a^2}{b^2}.\frac{c^2}{a^2}}=2|\frac{c}{b}|\geq \frac{2c}{b}\)
\(\frac{b^2}{c^2}+\frac{c^2}{a^2}\geq 2\sqrt{\frac{b^2}{c^2}.\frac{c^2}{a^2}}=2|\frac{b}{a}|\geq \frac{2b}{a}\)
Cộng theo vế:
\(2\left(\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}\right)\geq 2\left(\frac{c}{b}+\frac{b}{a}+\frac{a}{c}\right)\)
\(\Leftrightarrow \frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}\geq \frac{c}{b}+\frac{b}{a}+\frac{a}{c}\) (đpcm)
Dấu bằng xảy ra khi \(a=b=c\)
a)\(BDT\Leftrightarrow\dfrac{\left(a+b+c\right)\left(x+y+z\right)}{9}\le\dfrac{ax+by+cz}{3}\)
\(\Leftrightarrow\left(a+b+c\right)\left(x+y+z\right)\le3\left(ax+by+cz\right)\)
\(\Leftrightarrow ax+ay+az+bx+by+bz+cx+cy+cz\le3\left(ax+by+cz\right)\)
\(\Leftrightarrow2ax+2by+2cz-ay-az-bx-bz-cy-cx\ge0\)
\(\Leftrightarrow\left(ax-ay-bx+by\right)+\left(by-bz-cy+cz\right)+\left(cz-cx-az+ax\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)\left(x-y\right)+\left(b-c\right)\left(y-z\right)+\left(c-a\right)\left(z-x\right)\ge0\)
Đây là BĐT Chebyshev mình nghĩ phải có thêm điều kiện \(x\ge y\ge z\)
b)Nhân VP áp dụng Cauchy-Schwarz
c)Xem câu hỏi
Áp dụng BĐT AM-GM ta có:
\(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}\ge2\sqrt{\dfrac{a^2}{b^2}\cdot\dfrac{b^2}{c^2}}=2\sqrt{\dfrac{a^2}{c^2}}=\dfrac{2a}{c}\)
\(\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\ge2\sqrt{\dfrac{b^2}{c^2}\cdot\dfrac{c^2}{a^2}}=2\sqrt{\dfrac{b^2}{a^2}}=\dfrac{2b}{a}\)
\(\dfrac{c^2}{a^2}+\dfrac{a^2}{b^2}\ge2\sqrt{\dfrac{c^2}{a^2}\cdot\dfrac{a^2}{b^2}}=2\sqrt{\dfrac{c^2}{b^2}}=\dfrac{2c}{b}\)
Cộng theo vế 3 BĐT trên ta có:
\(\dfrac{2a^2}{b}+\dfrac{2b^2}{c}+\dfrac{2c^2}{a}\ge\dfrac{2a}{c}+\dfrac{2b}{a}+\dfrac{2c}{b}\)
\(\Leftrightarrow2\left(\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{a}\right)\ge2\left(\dfrac{a}{c}+\dfrac{b}{a}+\dfrac{c}{b}\right)\)
\(\Leftrightarrow\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{a}\ge\dfrac{a}{c}+\dfrac{b}{a}+\dfrac{c}{b}\)
Đẳng thức xảy ra khi \(a=b=c\)
CMR: \(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\ge\dfrac{c}{b}+\dfrac{b}{a}+\dfrac{a}{c}\)
Áp dụng bđt AM - GM ta có :
\(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}\ge2\sqrt{\dfrac{a^2}{b^2}.\dfrac{b^2}{c^2}}=2\left|\dfrac{a}{c}\right|\ge2\dfrac{a}{c}\)(1)
\(\dfrac{a^2}{b^2}+\dfrac{c^2}{a^2}\ge2\sqrt{\dfrac{a^2}{b^2}.\dfrac{c^2}{a^2}}=2\left|\dfrac{c}{b}\right|\ge2\dfrac{c}{b}\)(2)
\(\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\ge2\sqrt{\dfrac{b^2}{c^2}.\dfrac{c^2}{a^2}}=2\left|\dfrac{b}{a}\right|\ge2\dfrac{b}{a}\)(3)
Cộng vế với vế của (1);(2);(3) ta được :
\(2\left(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\right)\ge2\left(\dfrac{c}{b}+\dfrac{b}{a}+\dfrac{a}{c}\right)\)
\(\Rightarrow\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\ge\dfrac{c}{b}+\dfrac{b}{a}+\dfrac{a}{c}\)(đpcm)
Dấu "=" xảy ra \(\Leftrightarrow a=b=c\)
áp dụngBĐt cô si cho 2 số ta có
\(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}\ge2\sqrt{\dfrac{a^2}{b^2}.\dfrac{b^2}{c^2}}=2\dfrac{a}{c}\)
tt ta có
\(\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\ge2\dfrac{b}{a}\); \(\dfrac{b^2}{a^2}+\dfrac{a^2}{c^2}\ge2\dfrac{b}{c}\)
cộng các BĐT trên ta có
\(2\left(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\right)\ge2\left(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\right)\)
⇔ \(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\ge\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\) (đpcm)