Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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
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)
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 :
\(\dfrac{a}{b^2}+\dfrac{b}{c^2}+\dfrac{c}{a^2}+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\) ≥ \(2\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
⇔\(\dfrac{a}{b^2}+\dfrac{b}{c^2}+\dfrac{c}{a^2}\) ≥ \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\)
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 \(\Leftrightarrow a=b=c\)
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 Cauchy ta có
\(\dfrac{a^2}{b+c}+\dfrac{b+c}{4}\ge a\)
\(\dfrac{b^2}{a+c}+\dfrac{a+c}{4}\ge b\)
\(\dfrac{c^2}{a+b}+\dfrac{a+b}{4}\ge c\)
\(\Rightarrow\dfrac{a^2}{b+c}+\dfrac{b^2}{a+c}+\dfrac{c^2}{a+b}+\dfrac{a+b+c}{2}\ge a+b+c\)
\(\Rightarrow\dfrac{a^2}{b+c}+\dfrac{b^2}{a+c}+\dfrac{c^2}{a+b}\ge\dfrac{a+b+c}{2}\)
Dấu bằng xảy ra khi a=b=c
Làm tắt vài chỗ thông cảm
Câu b,
Ta có BĐT Cauchy \(a^2+b^2\ge2ab\)
\(\Rightarrow\left(a+b\right)^2\ge4ab\)
\(\Rightarrow ab\le\dfrac{\left(a+b\right)^2}{4}\)
\(\Rightarrow\dfrac{ab}{a+b}\le\dfrac{\left(a+b\right)^2}{4\left(a+b\right)}=\dfrac{a+b}{4}\)
Tương tự \(\dfrac{bc}{b+c}\le\dfrac{b+c}{4}\)
\(\dfrac{ac}{a+c}\le\dfrac{a+c}{4}\)
Cộng theo vế ta đc \(VT\le\dfrac{2\left(a+b+c\right)}{4}=\dfrac{a+b+c}{2}\)
Dấu bằng xảy ra khi a=b=c
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 $a=b=1$
e)
\(\dfrac{a^2+b^2+c^2}{3}\ge\left(\dfrac{a+b+c}{3}\right)^2\)
\(\Leftrightarrow3\left(a^2+b^2+c^2\right)\ge a^2+b^2+c^2+2\left(ab+bc+ca\right)\)
\(\Leftrightarrow2\left(a^2+b^2+c^2\right)\ge2\left(ab+bc+ac\right)\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2ac-2bc\ge0\)
\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(a^2-2ac+c^2\right)+\left(b^2-2bc+c^2\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(a-c\right)^2+\left(b-c\right)^2\ge0\) ( luôn đúng)
=> ĐPCM
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 .