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Không mất tính tổng quát, chuẩn hóa a + b + c = 1
Khi đó, ta cần chứng minh: \(\frac{\left(a+1\right)^2}{2a^2+\left(1-a\right)^2}+\frac{\left(b+1\right)^2}{2b^2+\left(1-b\right)^2}+\frac{\left(c+1\right)^2}{2c^2+\left(1-c\right)^2}\le8\)
Xét bất đẳng thức phụ: \(\frac{\left(x+1\right)^2}{2x^2+\left(1-x\right)^2}\le4x+\frac{4}{3}\)(*)
Thật vậy: (*)\(\Leftrightarrow\frac{\left(3x-1\right)^2\left(4x+1\right)}{2x^2+\left(1-x\right)^2}\ge0\)*đúng*
Áp dụng, ta được: \(\frac{\left(a+1\right)^2}{2a^2+\left(1-a\right)^2}+\frac{\left(b+1\right)^2}{2b^2+\left(1-b\right)^2}+\frac{\left(c+1\right)^2}{2c^2+\left(1-c\right)^2}\)\(\le4\left(a+b+c\right)+4=4.1+4=8\)
Vậy bất đẳng thức được chứng minh
Đẳng thức xảy ra khi a = b = c
Chuẩn hóa ta có : \(a+b+c=3\)
=> \(\frac{\left(2a+b+c\right)^2}{2a^2+\left(b+c\right)^2}=\frac{\left(a+3\right)^2}{2a^2+\left(3-a\right)^2}=\frac{a^2+6a+9}{3\left(a^2-2a+3\right)}\)
Xét\(\frac{a^2+6a+9}{3\left(a^2-2a+3\right)}\le\frac{4}{3}a+\frac{4}{3}\)
<=> \(a^2+6a+9\le4\left(a+1\right)\left(a^2-2a+3\right)\)
<=> \(4a^3-5a^2-2a+3\ge0\)
<=> \(\left(a-1\right)^2\left(4a+3\right)\ge0\)luôn đúng
Khi đó
\(VT\le\frac{4}{3}\left(a+b+c\right)+4=\frac{4}{3}.3+4=8\)(ĐPCM)
Dấu bằng xảy ra khi a=b=c
Bài 1:
Áp dụng BĐT Bunhiacopxky:
\(\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)[a(b+c)+b(c+a)+c(a+b)]\geq (a+b+c)^2\)
\(\Rightarrow \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\geq \frac{(a+b+c)^2}{2(ab+bc+ac)}\)$(*)$
Áp dụng BĐT AM-GM dễ thấy: $a^2+b^2+c^2\geq ab+bc+ac$
$\Rightarrow (a+b+c)^2\geq 3(ab+bc+ac)\Rightarrow ab+bc+ac\leq \frac{(a+b+c)^2}{3}(**)$
Từ $(*); (**)\Rightarrow \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\geq \frac{(a+b+c)^2}{2.\frac{(a+b+c)^2}{3}}=\frac{3}{2}$ (đpcm)
Dấu "=" xảy ra khi $a=b=c$
Bài 2:
Áp dụng BĐT AM-GM:
\(\frac{a^3}{b(2c+a)}+\frac{b}{3}+\frac{2c+a}{9}\geq 3\sqrt[3]{\frac{a^3}{b(2c+a)}.\frac{b}{3}.\frac{2c+a}{9}}=a\)
\(\frac{b^3}{c(2a+b)}+\frac{c}{3}+\frac{2a+b}{9}\geq b\)
\(\frac{c^3}{a(2b+c)}+\frac{a}{3}+\frac{2b+c}{9}\ge c\)
Cộng theo vế và thu gọn ta có:
\(\frac{a^3}{b(2c+a)}+\frac{b^3}{c(2a+b)}+\frac{c^3}{a(2b+c)}\geq \frac{a+b+c}{3}=\frac{3}{3}=1\) (đpcm)
Dấu "=" xảy ra khi $a=b=c=1$
\(\frac{a^2}{a^2+c^2}+\frac{b^2}{b^2+c^2}\ge\frac{\left(a+b\right)^2}{a^2+b^2+2c^2}\)
\(\frac{b^2}{b^2+a^2}+\frac{c^2}{c^2+a^2}\ge\frac{\left(b+c\right)^2}{b^2+c^2+2a^2}\)
\(\frac{c^2}{c^2+b^2}+\frac{a^2}{a^2+b^2}\ge\frac{\left(c+a\right)^2}{c^2+a^2+2b^2}\)
\(\Rightarrow VT\le\frac{a^2+c^2}{a^2+c^2}+\frac{b^2+c^2}{b^2+c^2}+\frac{a^2+b^2}{a^2+b^2}=1+1+1=3\)
Áp dụng BĐT Cauchy-Schwarz: \(\frac{a^2}{x}+\frac{b^2}{y}\ge\frac{\left(a+b\right)^2}{x+y}\)
Ta có \(\frac{\left(a+b\right)^2}{a^2+b^2+2c^2}=\frac{\left(a+b\right)^2}{a^2+c^2+b^2+c^2}\le\frac{a^2}{a^2+c^2}+\frac{b^2}{b^2+c^2}\)
Tương tự ta có:
\(\frac{\left(b+c\right)^2}{b^2+c^2+2a^2}\le\frac{b^2}{a^2+b^2}+\frac{c^2}{a^2+c^2}\) ; \(\frac{\left(c+a\right)^2}{c^2+a^2+2b^2}\le\frac{c^2}{b^2+c^2}+\frac{a^2}{a^2+b^2}\)
Cộng vế với vế:
\(\frac{\left(a+b\right)^2}{a^2+b^2+2c^2}+\frac{\left(b+c\right)^2}{b^2+c^2+2a^2}+\frac{\left(c+a\right)^2}{c^2+a^2+2b^2}\le\frac{a^2+c^2}{a^2+c^2}+\frac{b^2+c^2}{b^2+c^2}+\frac{a^2+b^2}{a^2+b^2}=3\)
Dấu "=" xảy ra khi \(a=b=c\)
//Bạn chép đề sai, vế phải là số 3 chứ ko phải 1
\(\frac{bc}{a^2\left(b+c\right)}+\frac{ca}{b^2\left(c+a\right)}+\frac{ab}{c^2\left(a+b\right)}\ge\frac{1}{2a}+\frac{1}{2b}+\frac{1}{2c}\)
\(\Rightarrow\frac{bc}{a^2\left(b+c\right)}+\frac{b+c}{4bc}\ge2\sqrt{\frac{bc}{a^2\left(b+c\right)}\cdot\frac{b+c}{4bc}}=\frac{1}{a}\)
\(\Rightarrow\frac{ca}{b^2\left(c+a\right)}+\frac{c+a}{4ca}\ge2\sqrt{\frac{ca}{b^2\left(c+a\right)}\cdot\frac{c+a}{4ca}}=\frac{1}{b}\)
\(\Rightarrow\frac{ab}{c^2\left(a+b\right)}+\frac{a+b}{4ab}\ge2\sqrt{\frac{ab}{c^2\left(a+b\right)}\cdot\frac{a+b}{4ab}}=\frac{1}{c}\)
Cộng theo vế các bất đẳng thức trên ta được:
\(\frac{bc}{a^2\left(b+c\right)}+\frac{ca}{b^2\left(c+a\right)}+\frac{ab}{c^2\left(a+b\right)}+\frac{b+c}{4bc}+\frac{c+a}{4ca}+\frac{a+b}{4ab}\ge\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)
Mà\(\frac{b+c}{4bc}+\frac{c+a}{4ca}+\frac{a+b}{4ab}=\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)nên:
\(\frac{bc}{a^2\left(b+c\right)}+\frac{ca}{b^2\left(c+a\right)}+\frac{ab}{c^2\left(a+b\right)}\ge\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
hay\(\frac{bc}{a^2\left(b+c\right)}+\frac{ca}{b^2\left(c+a\right)}+\frac{ab}{c^2\left(a+b\right)}\ge\frac{1}{2a}+\frac{1}{2b}+\frac{1}{2c}\)
Bất đẳng thức xảy ra khi \(a=b=c\)
1)
\(2a+\frac{4}{a}+\frac{16}{a+2}=\left(a+\frac{4}{a}\right)+\left[\left(a+2\right)+\frac{16}{a+2}\right]-2\ge4+8-2=10\)
Dấu "=" xảy ra khi a=2
2)
\(\hept{\begin{cases}\sqrt{a\left(1-4a\right)}=\frac{1}{2}\sqrt{4a\left(1-4a\right)}\le\frac{1}{2}\cdot\frac{4a+1-4a}{2}=\frac{1}{4}\\\sqrt{b\left(1-4b\right)}=\frac{1}{2}\sqrt{4\left(1-4a\right)}\le\frac{1}{2}\cdot\frac{4b+1-4b}{2}=\frac{1}{4}\\\sqrt{c\left(1-4c\right)}=\frac{1}{2}\sqrt{4c\left(1-4c\right)}\le\frac{1}{2}\cdot\frac{4c+1-4c}{2}=\frac{1}{4}\end{cases}}\)
\(\Rightarrow\sqrt{a\left(1-4a\right)}+\sqrt{b\left(1-4b\right)}+\sqrt{c\left(1-4c\right)}\le\frac{3}{4}\)
Dấu "=" xảy ra khi \(a=b=c=\frac{1}{8}\)