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\(\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\ge\frac{a+b+c}{2}\)
\(\Leftrightarrow\frac{a^2}{b+c}+a+\frac{b^2}{a+c}+b+\frac{c^2}{a+b}+c\ge\frac{a+b+c}{2}+a+b+c\)
\(\Leftrightarrow a\left(\frac{a}{b+c}+1\right)+b\left(\frac{b}{a+c}+1\right)+c\left(\frac{c}{a+b}+1\right)\ge\frac{3}{2}\left(a+b+c\right)\)
\(\Leftrightarrow a\left(\frac{a+b+c}{b+c}\right)+b\left(\frac{a+b+c}{c+a}\right)+c\left(\frac{a+b+c}{a+b}\right)\ge\frac{3}{2}\left(a+b+c\right)\)
\(\Leftrightarrow\left(a+b+c\right)\frac{a}{b+c}+\left(a+b+c\right)\frac{b}{c+a}+\left(a+b+c\right)\frac{c}{a+b}\ge\frac{3}{2}\left(a+b+c\right)\)
\(\Leftrightarrow\left(a+b+c\right)\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)\ge\frac{3}{2}\left(a+b+c\right)\)
\(\Leftrightarrow\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge\frac{3}{2}\)
\(\Leftrightarrow\frac{a}{b+c}+1+\frac{b}{c+a}+1+\frac{c}{a+b}+1\ge\frac{3}{2}+3\)
\(\Leftrightarrow\frac{a+b+c}{b+c}+\frac{a+b+c}{c+a}+\frac{a+b+c}{a+b}\ge\frac{9}{2}\)
\(\Leftrightarrow\left(a+b+c\right)\left(\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}\right)\ge\frac{9}{2}\)
\(\Leftrightarrow2\left(a+b+c\right)\left(\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}\right)\ge9\)
\(\Leftrightarrow\left(2a+2b+2c\right)\left(\frac{1}{b+c}+\frac{1}{a+c}+\frac{1}{a+b}\right)\ge9\)
\(\Leftrightarrow\left(b+c+c+a+a+b\right)\left(\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}\right)\ge9\)
Áp dụng BĐT Cô - si
\(\Rightarrow\left\{\begin{matrix}b+c+c+a+a+b\ge3\sqrt[3]{\left(b+c\right)\left(c+a\right)\left(a+b\right)}\\\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}\ge3\sqrt[3]{\frac{1}{\left(b+c\right)\left(c+a\right)\left(a+b\right)}}\end{matrix}\right.\)
Nhân từng vế :
\(\Rightarrow\left(b+c+c+a+a+b\right)\left(\frac{1}{b+c}+\frac{1}{a+c}+\frac{1}{a+b}\right)\ge9\sqrt[3]{\left(b+c\right)\left(c+a\right)\left(a+b\right).\frac{1}{\left(b+c\right)\left(c+a\right)\left(a+b\right)}}\)
\(\Rightarrow\left(b+c+c+a+a+b\right)\left(\frac{1}{b+c}+\frac{1}{a+c}+\frac{1}{a+b}\right)\ge9\left(đpcm\right)\)
Vậy với a ,b ,c > 0 thì \(\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\ge\frac{a+b+c}{2}\)
Áp dụng bất đẳng thức cô-si cho các số thực không âm ta có:
\(\frac{a^2}{b+c}+\frac{b+c}{4}\ge2\sqrt{\frac{a^2}{b+c}\times\frac{b+c}{4}}=a\) (1)
\(\frac{b^2}{a+c}+\frac{a+c}{4}\ge2\sqrt{\frac{b^2}{a+c}\times\frac{a+c}{4}}=b\) (2)
\(\frac{c^2}{a+b}+\frac{a+b}{4}\ge2\sqrt{\frac{c^2}{a+b}\times\frac{a+b}{4}}=c\) (3)
Cộng (1),(2) và (3),vế theo vế ta được:
\(\frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{a+b}+\frac{a+b+c}{2}\ge a+b+c\)
\(\Rightarrow\frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{a+b}\ge\frac{a+b+c}{2}\) (đpcm)
Dấu "=" xảy ra khi :a=b=c
Vậy \(\frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{a+b}\ge\frac{a+b+c}{2}\) với a,b,c >0
mình nghĩ đề bài sai một chỗ :\(\frac{a^2}{b^2}\)chứ ko phải là \(\frac{a}{b^2}\)
Áp dụng bất đẳng thức Bu-nhi-a mở rộng, ta có:
\(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\)
\(\ge\frac{\left(a+b+c\right)^2}{a+b+c}\)
\(=a+b+c\)
Dấu "=" xảy ra khi a=b=c
Do a,b,c dương nên áp dụng cô-si cho 2 số dương \(\frac{a^2}{b}\)và\(b\)ta được
\(\frac{a^2}{b}+b\ge2\sqrt{\frac{a^2}{b}\cdot b}=2a\)
Tương tự
\(\frac{b^2}{c}+c\ge2b\): \(\frac{c^2}{a}+a\ge2c\)
\(\Rightarrow\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}+a+b+c\ge2a+2b+2c\)
\(\Rightarrow\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\ge a+b+c\)
bài này nhiều cách làm nhưng bn xem thử cách này nhé
Áp dụng bất đẳng thức Cô-si ta có:
\(\dfrac{a^2}{b^3}+\dfrac{1}{a}+\dfrac{1}{a}\ge\sqrt[3]{\dfrac{a^2}{b^3}.\dfrac{1}{a}.\dfrac{1}{a}}=\dfrac{3}{b}\)
\(\dfrac{c^2}{a^3}+\dfrac{1}{c}+\dfrac{1}{c}\ge\sqrt[3]{\dfrac{c^2}{a^3}.\dfrac{1}{c}.\dfrac{1}{c}}=\dfrac{3}{a}\)
\(\dfrac{c^2}{a^3}+\dfrac{1}{c}+\dfrac{1}{c}\ge\sqrt[3]{\dfrac{c^2}{a^3}.\dfrac{1}{c}.\dfrac{1}{c}}=\dfrac{3}{a}\)
Cộng theo vế ta được:
\(\dfrac{a^2}{b^3}+\dfrac{b^2}{c^3}+\dfrac{a^2}{a^3}+\dfrac{2}{a}+\dfrac{2}{b}+\dfrac{2}{c}\ge3\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
\(\Leftrightarrow\dfrac{a^2}{b^3}+\dfrac{b^2}{c^3}+\dfrac{c^2}{a^3}\ge\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\)
Áp dụng Svac
\(\Sigma\frac{a^3}{b+c}=\Sigma\frac{a^4}{ab+ac}\ge\frac{\left(a^2+b^2+c^2\right)^2}{2\left(ab+bc+ca\right)}\ge\frac{\left(a^2+b^2+c^2\right)^2}{2\left(a^2+b^2+c^2\right)}=\frac{1}{2}\left(a^2+b^2+c^2\right)\)
"=" tại a=b=c
E thử làm cách khác ạ:))
Không mất tính tổng quát,giả sử \(a\ge b\ge c\)
\(\Rightarrow\hept{\begin{cases}a^2\ge b^2\ge c^2\\\frac{a}{b+c}\ge\frac{b}{a+c}\ge\frac{c}{a+b}\end{cases}}\)
Áp dụng BĐT Trebysev ta có:
\(a^2\cdot\frac{a}{b+c}+b^2\cdot\frac{b}{a+c}+c^2\cdot\frac{c}{a+b}\ge\frac{a^2+b^2+c^2}{3}\cdot\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)\)
\(\ge\frac{a^2+b^2+c^2}{3}\cdot\frac{3}{2}\left(nesbitt\right)\)
\(=\frac{a^2+b^2+c^2}{2}\)
Dấu "=" xảy ra khi \(a=b=c\)
\(\frac{a^2}{b^3}+\frac{b^2}{c^3}+\frac{c^2}{a^3}=\frac{1}{b}+\frac{1}{c}+\frac{1}{a}\)
=> \(\frac{a^2}{b^3}+\frac{b^2}{c^3}+\frac{c^2}{a^3}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)
Kết hợp Mincôpxki và C-S:
\(VT\ge\sqrt{\left(\frac{3}{a+b}+\frac{3}{b+c}+\frac{3}{a+c}\right)^2+\left(a+b+c\right)^2}\)
\(VT\ge\sqrt{\left(\frac{27}{2\left(a+b+c\right)}\right)^2+\left(a+b+c\right)^2}=\sqrt{\frac{405}{4\left(a+b+c\right)^2}+\frac{81}{\left(a+b+c\right)^2}+\left(a+b+c\right)^2}\)
\(VT\ge\sqrt{\frac{405}{12\left(a^2+b^2+c^2\right)}+2\sqrt{\frac{81\left(a+b+c\right)^2}{\left(a+b+c\right)^2}}}=\sqrt{\frac{405}{12.3}+18}=\frac{3\sqrt{13}}{2}\) (đpcm)
Dấu "=" xảy ra khi \(a=b=c=1\)
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