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Ta đi chứng minh BĐT : \(a^2+b^2+c^2\ge2\left(bc+ac-ab\right)\)
\(\Leftrightarrow\) \(a^2+b^2+c^2+2ab-2bc-2ac\ge0\)
\(\Leftrightarrow\) \(\left(a+b-c\right)^2\ge0\) luôn đúng.
\(\Rightarrow2\left(bc+ac-ab\right)\le\dfrac{5}{3}\)
\(\Leftrightarrow bc+ac-ab\le\dfrac{5}{6}< 1\)
\(\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}-\dfrac{1}{c}< \dfrac{1}{abc}\)
Xét: \(\dfrac{a+1}{b^2+1}+\dfrac{b+1}{c^2+1}+\dfrac{c+1}{a^2+1}\)
\(\Leftrightarrow\dfrac{a}{b^2+1}+\dfrac{b}{c^2+1}+\dfrac{c}{a^2+1}+\dfrac{1}{a^2+1}+\dfrac{1}{b^2+1}+\dfrac{1}{c^2+1}\)
\(\Leftrightarrow a-\dfrac{ab^2}{b^2+1}+b-\dfrac{bc^2}{c^2+1}+c-\dfrac{ca^2}{a^2+1}+1-\dfrac{a^2}{a^2+1}+1-\dfrac{b^2}{b^2+1}+1-\dfrac{c^2}{c^2+1}\)
\(\Leftrightarrow3-\left(\dfrac{ab^2}{b^2+1}+\dfrac{bc^2}{c^2+1}+\dfrac{ca^2}{a^2+1}\right)+3-\left(\dfrac{a^2}{a^2+1}+\dfrac{b^2}{b^2+1}+\dfrac{c^2}{c^2+1}\right)\)
Xét \(3-\left(\dfrac{ab^2}{b^2+1}+\dfrac{bc^2}{c^2+1}+\dfrac{ca^2}{a^2+1}\right)\)
Áp dụng bất đẳng thức Cauchy cho 2 bộ số thực không âm
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{ab^2}{b^2+1}\le\dfrac{ab^2}{2b}=\dfrac{ab}{2}\\\dfrac{bc^2}{c^2+1}\le\dfrac{bc^2}{2c}=\dfrac{bc}{2}\\\dfrac{ca^2}{a^2+1}\le\dfrac{ca^2}{2a}=\dfrac{ca}{2}\end{matrix}\right.\)
\(\Rightarrow\dfrac{ab^2}{b^2+1}+\dfrac{bc^2}{c^2+1}+\dfrac{ca^2}{a^2+1}\le\dfrac{ab+bc+ca}{2}\)
\(\Rightarrow3-\left(\dfrac{ab^2}{b^2+1}+\dfrac{bc^2}{c^2+1}+\dfrac{ca^2}{a^2+1}\right)\ge3-\dfrac{ab+bc+ca}{2}\) ( 1 )
Theo hệ quả của bất đẳng thức Cauchy ta có
\(\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\)
\(\Rightarrow\dfrac{3}{2}\le3-\dfrac{ab+bc+ca}{2}\) ( 2 )
Từ ( 1 ) và ( 2 )
\(\Rightarrow3-\left(\dfrac{ab^2}{b^2+1}+\dfrac{bc^2}{c^2+1}+\dfrac{ca^2}{a^2+1}\right)\ge\dfrac{3}{2}\) ( 3 )
Xét \(3-\left(\dfrac{a^2}{a^2+1}+\dfrac{b^2}{b^2+1}+\dfrac{c^2}{c^2+1}\right)\)
Áp dụng bất đẳng thức Cauchy cho 2 bộ số thực không âm
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{a^2}{a^2+1}\le\dfrac{a^2}{2a}=\dfrac{a}{2}\\\dfrac{b^2}{b^2+1}\le\dfrac{b^2}{2b}=\dfrac{b}{2}\\\dfrac{c^2}{c^2+1}\le\dfrac{c^2}{2c}=\dfrac{c}{2}\end{matrix}\right.\)
\(\Rightarrow\dfrac{a^2}{a^2+1}+\dfrac{b^2}{b^2+1}+\dfrac{c^2}{c^2+1}\le\dfrac{a+b+c}{2}=\dfrac{3}{2}\)
\(\Rightarrow3-\left(\dfrac{a^2}{a^2+1}+\dfrac{b^2}{b^2+1}+\dfrac{c^2}{c^2+1}\right)\ge3-\dfrac{3}{2}=\dfrac{3}{2}\) ( 4 )
Từ ( 3 ) và ( 4 ) cộng theo từng vế
\(\Rightarrow VT\ge\dfrac{3}{2}+\dfrac{3}{2}=3\)
\(\Leftrightarrow\dfrac{a+1}{b^2+1}+\dfrac{b+1}{c^2+1}+\dfrac{c+1}{a^2+1}\ge3\)
\(\Rightarrow\) ( đpcm )
\(\dfrac{a}{4b^2+1}+\dfrac{b}{4a^2+1}=\dfrac{a\left(4b^2+1\right)}{4b^2+1}-\dfrac{4ab^2}{4b^2+1}+\dfrac{b\left(4a^2+1\right)}{4a^2+1}-\dfrac{4a^2b}{4b^2+1}\)
\(\ge a-\dfrac{4ab^2}{4b}+b-\dfrac{4a^2b}{4a}\) (bđt Cô-si)
=a-ab+b-ab=a+b-2ab=4ab-2ab=2ab
Lại có a+b=4ab \(\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}=4\ge\dfrac{2}{2\sqrt{ab}}\Rightarrow4\sqrt{ab}\ge2\Rightarrow ab\ge\dfrac{1}{4}\)
\(\Rightarrow2ab\ge\dfrac{1}{2}\Rightarrow\dfrac{a}{4b^2+1}+\dfrac{b}{4a^2+1}\ge\dfrac{1}{2}\)
Dấu ''='' xảy ra khi \(a=b=\dfrac{1}{2}\)
\(\dfrac{a}{4b^2+1}+\dfrac{b}{4a^2+1}\ge\dfrac{1}{2}\)
\(\Leftrightarrow a-\dfrac{a}{4b^2+1}+b-\dfrac{b}{4a^2+1}\le a+b-\dfrac{1}{2}\)
\(\Rightarrow\dfrac{4ab^2}{4b^2+1}+\dfrac{4ba^2}{4a^2+1}\le4ab-\dfrac{1}{2}\)
\(\sum\dfrac{4ab^2}{4b^2+1}\le^{CS}2ab\)
\(\Rightarrow CM:2ab\le4ab-\dfrac{1}{2}\Leftrightarrow ab\ge\dfrac{1}{4}\)
Từ GT \(\Rightarrow4ab=a+b\ge2\sqrt{ab}\Leftrightarrow ab\ge\dfrac{1}{4}\)
\(\Rightarrow dpcm\)
Câu 3. Dự đoán dấu "=" khi \(a=b=c=\frac{1}{\sqrt{3}}\)
Dùng phương pháp chọn điểm rơi thôi :)
LG
Áp dụng bđt Cô-si được \(a^2+b^2+c^2\ge3\sqrt[3]{a^2b^2c^2}\)
\(\Rightarrow1\ge3\sqrt[3]{a^2b^2c^2}\)
\(\Rightarrow\frac{1}{3}\ge\sqrt[3]{a^2b^2c^2}\)
\(\Rightarrow\frac{1}{27}\ge a^2b^2c^2\)
\(\Rightarrow\frac{1}{\sqrt{27}}\ge abc\)
Khi đó :\(B=a+b+c+\frac{1}{abc}\)
\(=a+b+c+\frac{1}{9abc}+\frac{8}{9abc}\)
\(\ge4\sqrt[4]{abc.\frac{1}{9abc}}+\frac{8}{9.\frac{1}{\sqrt{27}}}\)
\(=4\sqrt[4]{\frac{1}{9}}+\frac{8\sqrt{27}}{9}=\frac{4}{\sqrt[4]{9}}+\frac{8}{\sqrt{3}}=\frac{4}{\sqrt{3}}+\frac{8}{\sqrt{3}}=\frac{12}{\sqrt{3}}=4\sqrt{3}\)
Dấu "=" \(\Leftrightarrow a=b=c=\frac{1}{\sqrt{3}}\)
Vậy .........
2, \(A=\frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{a+b}\)
\(A=\frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{a+b}\)
\(A=\left[\frac{a^2}{b+c}+\frac{\left(b+c\right)}{4}\right]+\left[\frac{b^2}{a+c}+\frac{\left(a+c\right)}{4}\right]+\left[\frac{c^2}{a+b}+\frac{\left(a+b\right)}{4}\right]-\frac{\left(a+b+c\right)}{2}\)
Áp dụng BĐT AM-GM ta có:
\(A\ge2.\sqrt{\frac{a^2}{4}}+2.\sqrt{\frac{b^2}{4}}+2.\sqrt{\frac{c^2}{4}}-\frac{\left(a+b+c\right)}{2}\)
\(A\ge a+b+c-\frac{6}{2}\)
\(A\ge6-3\)
\(A\ge3\)
Dấu " = " xảy ra \(\Leftrightarrow\)\(\frac{a^2}{b+c}=\frac{b+c}{4}\Leftrightarrow4a^2=\left(b+c\right)^2\Leftrightarrow2a=b+c\)(1)
\(\frac{b^2}{a+c}=\frac{a+c}{4}\Leftrightarrow4b^2=\left(a+c\right)^2\Leftrightarrow2b=a+c\)(2)
\(\frac{c^2}{a+b}=\frac{a+b}{4}\Leftrightarrow4c^2=\left(a+b\right)^2\Leftrightarrow2c=a+b\)(3)
Lấy \(\left(1\right)-\left(3\right)\)ta có:
\(2a-2c=c+b-a-b=c-a\)
\(\Rightarrow2a-2c-c+a=0\)
\(\Leftrightarrow3.\left(a-c\right)=0\)
\(\Leftrightarrow a-c=0\Leftrightarrow a=c\)
Chứng minh tương tự ta có: \(\hept{\begin{cases}b=c\\a=b\end{cases}}\)
\(\Rightarrow a=b=c=2\)
Vậy \(A_{min}=3\Leftrightarrow a=b=c=2\)
\(\frac{1}{a^2+b^2+2}+\frac{1}{c^2+b^2+2}+\frac{1}{a^2+c^2+2}\le\frac{3}{4}\)
\(\Leftrightarrow\frac{a^2+b^2}{a^2+b^2+2}+\frac{b^2+c^2}{b^2+c^2+2}+\frac{c^2+a^2}{c^2+a^2+2}\ge\frac{3}{2}\)
Áp dụng BĐT Cauchy-Schwarz ta có:
\(VT\ge\frac{\left(\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{c^2+a^2}\right)^2}{2\left(a^2+b^2+c^2\right)+6}\)
\(\ge\frac{\sqrt{3\left(a^2b^2+b^2c^2+c^2a^2\right)}+2\left(a^2+b^2+c^2\right)}{a^2+b^2+c^2}\)
\(\ge\frac{2\left(a^2+b^2+c^2\right)+ab+bc+ca}{a^2+b^2+c^2}\)
Cần chứng minh \(\frac{2\left(a^2+b^2+c^2\right)+ab+bc+ca}{a^2+b^2+c^2}\ge\frac{3}{2}\)
\(\Leftrightarrow\left(a+b+c\right)^2\ge0\) *luôn đúng*
Áp dụng bất đẳng thức Cauchy-Schwarz ta có:
\(\dfrac{1}{2a^2+b^2}=\dfrac{1}{a^2+a^2+b^2}\le\dfrac{1}{9}\left(\dfrac{1}{a^2}+\dfrac{1}{a^2}+\dfrac{1}{b^2}\right)\)
\(\left\{{}\begin{matrix}\dfrac{1}{2b^2+c^2}\le\dfrac{1}{9}\left(\dfrac{1}{b^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)\\\dfrac{1}{2c^2+a^2}\le\dfrac{1}{9}\left(\dfrac{1}{c^2}+\dfrac{1}{c^2}+\dfrac{1}{a^2}\right)\end{matrix}\right.\)
Cộng theo vế:
\(L\le\dfrac{1}{9}\left(\dfrac{3}{a^2}+\dfrac{3}{b^2}+\dfrac{3}{c^2}\right)=\dfrac{1}{9}\left[3\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)\right]=\dfrac{1}{9}\)