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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 Cô - Si , ta có :
\(\dfrac{a}{b^2}+\dfrac{1}{a}\) ≥ \(2\sqrt{\dfrac{a}{b^2}.\dfrac{1}{a}}=2.\dfrac{1}{b}\left(a,b>0\right)\left(1\right)\)
\(\dfrac{b}{c^2}+\dfrac{1}{b}\text{ ≥ }2\sqrt{\dfrac{b}{c^2}.\dfrac{1}{b}}=2.\dfrac{1}{c}\left(b,c>0\right)\left(2\right)\)
\(\dfrac{c}{a^2}+\dfrac{1}{c}\text{≥}2\sqrt{\dfrac{c}{a^2}.\dfrac{1}{c}}=2.\dfrac{1}{a}\left(a,c>0\right)\left(3\right)\)
Từ ( 1 ; 2 ; 3) Ta có :
\(\dfrac{a}{b^2}+\dfrac{b}{c^2}+\dfrac{c}{a^2}+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\) ≥ \(2\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
⇔\(\dfrac{a}{b^2}+\dfrac{b}{c^2}+\dfrac{c}{a^2}\) ≥ \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\)
ko xoắn 1 dòng thôi
\(\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}\ge\dfrac{\left(a+b+c\right)^2}{b+c+c+a+a+b}=\dfrac{\left(a+b+c\right)^2}{2\left(a+b+c\right)}=\dfrac{a+b+c}{2}\)
Xét:
\(\dfrac{a^2}{b^2+c^2}-\dfrac{a}{b+c}=\dfrac{a\left(ab+ac-b^2-c^2\right)}{\left(b^2+c^2\right)\left(b+c\right)}=\dfrac{ab\left(a-b\right)+ac\left(a-c\right)}{\left(b^2+c^2\right)\left(b+c\right)}\left(1\right)\)
Tương tự:
\(\dfrac{b^2}{c^2+a^2}-\dfrac{b}{c+a}=\dfrac{bc\left(b-c\right)+ba\left(b-a\right)}{\left(c^2+a^2\right)\left(c+a\right)}\) (2)
\(\dfrac{c^2}{a^2+b^2}-\dfrac{c}{a+b}=\dfrac{ca\left(c-a\right)+cb\left(c-b\right)}{\left(a^2+b^2\right)\left(a+b\right)}\) (3)
Cộng từng vế (1)(2)(3) ta được:
\(\left(\dfrac{a^2}{b^2+c^2}+\dfrac{b^2}{c^2+a^2}+\dfrac{c^2}{a^2+b^2}\right)-\left(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\right)\)
\(=ab\left(a-b\right)\left[\dfrac{1}{\left(b^2+c^2\right)\left(b+c\right)}-\dfrac{1}{\left(a^2+c^2\right)\left(a+c\right)}\right]+ac\left(a-c\right)\left[\dfrac{1}{\left(b^2+c^2\right)\left(b+c\right)}-\dfrac{1}{\left(a^2+b^2\right)\left(a+b\right)}\right]+bc\left(b-c\right)\left[\dfrac{1}{\left(a^2+c^2\right)\left(a+c\right)}-\dfrac{1}{\left(a^2+b^2\right)\left(a+b\right)}\right]\) => ĐPCM
Bài 1:
Vì $a,b,c$ là 3 cạnh tam giác nên \(b+c-a; c+a-b; a+b-c>0\)
Áp dụng BĐT AM-GM cho các số dương:
\(\frac{a^2}{b+c-a}+(b+c-a)\geq 2\sqrt{a^2}=2a\)
\(\frac{b^2}{a+c-b}+(a+c-b)\geq 2\sqrt{b^2}=2b\)
\(\frac{c^2}{a+b-c}+(a+b-c)\geq 2\sqrt{c^2}=2c\)
Cộng theo vế và rút gọn:
\(\Rightarrow \frac{a^2}{b+c-a}+\frac{b^2}{c+a-b}+\frac{c^2}{a+b-c}+a+b+c\geq 2(a+b+c)\)
\(\Rightarrow \frac{a^2}{b+c-a}+\frac{b^2}{c+a-b}+\frac{c^2}{a+b-c}\geq a+b+c\) (đpcm)
Dấu "=" xảy ra khi $a=b=c$
Bài 2:
Áp dụng BĐT AM-GM cho các số dương ta có:
\(ab+\frac{a}{b}\geq 2\sqrt{ab.\frac{a}{b}}=2a\)
\(ab+\frac{b}{a}\geq 2\sqrt{ab.\frac{b}{a}}=2b\)
\(\frac{a}{b}+\frac{b}{a}\geq 2\sqrt{\frac{a}{b}.\frac{b}{a}}=2\)
Cộng theo vế và rút gọn:
\(\Rightarrow 2(ab+\frac{a}{b}+\frac{b}{a})\geq 2(a+b+1)\)
\(\Rightarrow ab+\frac{a}{b}+\frac{b}{a}\geq a+b+1\)
Ta có đpcm
Dấu "=" xảy ra khi $a=b=1$
Cjo 3 số dương a,b,c . CMR
\(\dfrac{a^2}{b+c}+\dfrac{b^2}{a+c}+\dfrac{c^2}{a+b}\ge\dfrac{a+b+c}{2}\)
Do a,b,c dương
Để làm bài này bạn cần chứng minh BĐT sau\(\dfrac{x^2}{m}+\dfrac{y^2}{n}\ge\dfrac{\left(x+y\right)^2}{m+n}\)(m;n>0)
<=>(m+n)(nx2+my2)-mn(x+y)2\(\ge\)0
Mình làm tắt,rút gọn luôn
<=>n2x2-2mnxy+m2y2\(\ge\)0
<=>(nx-my)2\(\ge\)0
=>BĐT trên được chứng minh và dấu bằng xảy ra khi nx=my
Mở rộng cho 3 số \(\dfrac{x^2}{m}+\dfrac{y^2}{n}+\dfrac{z^2}{p}\ge\dfrac{\left(x+y+z\right)^2}{m+n+p}\)
Áp dụng BĐT trên ta được:
\(\dfrac{a^2}{b+c}+\dfrac{b^2}{a+c}+\dfrac{c^2}{a+b}\ge\dfrac{\left(a+b+c\right)^2}{2\left(a+b+c\right)}=\dfrac{a+b+c}{2}\)
Dấu = xảy ra khi a=b=c
a)Svac-so:
\(\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}\ge\dfrac{\left(a+b+c\right)^2}{b+c+c+a+a+b}=\dfrac{\left(a+b+c\right)^2}{2\left(a+b+c\right)}=\dfrac{a+b+c}{2\left(đpcm\right)}\)
b)\(\dfrac{1}{a^2+1}+\dfrac{1}{b^2+1}\ge\dfrac{2}{ab+1}\)
\(\Leftrightarrow\dfrac{1}{a^2+1}-\dfrac{1}{ab+1}+\dfrac{1}{b^2+1}-\dfrac{1}{ab+1}\ge0\)
\(\Leftrightarrow\dfrac{ab+1-a^2-1}{\left(a^2+1\right)\left(ab+1\right)}+\dfrac{ab+1-b^2-1}{\left(b^2+1\right)\left(ab+1\right)}\ge0\)
\(\Leftrightarrow\dfrac{a\left(b-a\right)}{\left(a^2+1\right)\left(ab+1\right)}+\dfrac{b\left(a-b\right)}{\left(b^2+1\right)\left(ab+1\right)}\ge0\)
\(\Leftrightarrow\left(a-b\right)\left(\dfrac{b}{\left(b^2+1\right)\left(ab+1\right)}-\dfrac{a}{\left(a^2+1\right)\left(ab+1\right)}\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)\left(\dfrac{b\left(a^2+1\right)-a\left(b^2+1\right)}{\left(a^2+1\right)\left(b^2+1\right)\left(ab+1\right)}\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)\left(\dfrac{a^2b+b-ab^2-a}{\left(a^2+1\right)\left(b^2+1\right)\left(ab+1\right)}\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)\left(\dfrac{ab\left(a-b\right)-\left(a-b\right)}{\left(a^2+1\right)\left(b^2+1\right)\left(ab+1\right)}\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)^2\cdot\dfrac{ab-1}{\left(a^2+1\right)\left(b^2+1\right)\left(ab+1\right)}\ge0\)(luôn đúng)
5. phân tích ra : \(1+\dfrac{a}{b}+\dfrac{b}{a}+1\)
áp dụng bđ cosy
\(\dfrac{a}{b}+\dfrac{b}{a}\ge2\sqrt{\dfrac{a}{b}.\dfrac{b}{a}}=2\)
=> đpcm
6. \(x^2-x+1=x^2-2.\dfrac{1}{2}.x+\dfrac{1}{4}+\dfrac{3}{4}=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\)
hay với mọi x thuộc R đều là nghiệm của bpt
7.áp dụng bđt cosy
\(a^4+b^4+c^4+d^4\ge2\sqrt{a^2.b^2.c^2.d^2}=4abcd\left(đpcm\right)\)
a)Áp dụng bđt Cô-si:
\(\dfrac{a}{b}+\dfrac{b}{a}-1+\dfrac{ab}{a^2-ab+b^2}=\dfrac{a^2+b^2-ab}{ab}+\dfrac{ab}{a^2-ab+b^2}\ge2\sqrt{\dfrac{a^2+b^2-ab}{ab}.\dfrac{ab}{a^2-ab+b^2}}=2\)
=>\(\dfrac{a}{b}+\dfrac{b}{a}+\dfrac{ab}{a^2-ab+b^2}\ge3\)
Dấu "=" xảy ra khi a=b=1
b) bđt sai rồi
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 \(\Leftrightarrow a=b=c\)