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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 \(A_{min}=3\Leftrightarrow a=b=c=2\)
Bài 2:
\(\sqrt{\dfrac{a}{b+c}}+\sqrt{\dfrac{b}{c+a}}+\sqrt{\dfrac{c}{a+b}}>2\)
Trước hết ta chứng minh \(\sqrt{\dfrac{a}{b+c}}\ge\dfrac{2a}{a+b+c}\)
Áp dụng BĐT AM-GM ta có:
\(\sqrt{a\left(b+c\right)}\le\dfrac{a+b+c}{2}\)\(\Rightarrow1\ge\dfrac{2\sqrt{a\left(b+c\right)}}{a+b+c}\)
\(\Rightarrow\sqrt{\dfrac{a}{b+c}}\ge\dfrac{2a}{a+b+c}\). Ta lại có:
\(\sqrt{\dfrac{a}{b+c}}=\dfrac{\sqrt{a}}{\sqrt{b+c}}=\dfrac{a}{\sqrt{a\left(b+c\right)}}\ge\dfrac{2a}{a+b+c}\)
Thiết lập các BĐT tương tự:
\(\sqrt{\dfrac{b}{c+a}}\ge\dfrac{2b}{a+b+c};\sqrt{\dfrac{c}{a+b}}\ge\dfrac{2c}{a+b+c}\)
Cộng theo vế 3 BĐT trên ta có:
\(VT\ge\dfrac{2a}{a+b+c}+\dfrac{2b}{a+b+c}+\dfrac{2c}{a+b+c}=\dfrac{2\left(a+b+c\right)}{a+b+c}\ge2\)
Dấu "=" không xảy ra nên ta có ĐPCM
Lưu ý: lần sau đăng từng bài 1 thôi nhé !
1) Áp dụng liên tiếp bđt \(\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{4}{x+y}\) với a;b là 2 số dương ta có:
\(\dfrac{1}{2a+b+c}=\dfrac{1}{\left(a+b\right)+\left(a+c\right)}\le\dfrac{\dfrac{1}{a+b}+\dfrac{1}{a+c}}{4}\)\(\le\dfrac{\dfrac{2}{a}+\dfrac{1}{b}+\dfrac{1}{c}}{16}\)
TT: \(\dfrac{1}{a+2b+c}\le\dfrac{\dfrac{2}{b}+\dfrac{1}{a}+\dfrac{1}{c}}{16}\)
\(\dfrac{1}{a+b+2c}\le\dfrac{\dfrac{2}{c}+\dfrac{1}{a}+\dfrac{1}{b}}{16}\)
Cộng vế với vế ta được:
\(\dfrac{1}{2a+b+c}+\dfrac{1}{a+2b+c}+\dfrac{1}{a+b+2c}\le\dfrac{1}{16}.\left(\dfrac{4}{a}+\dfrac{4}{b}+\dfrac{4}{c}\right)=1\left(đpcm\right)\)
Ta có BĐT \(3\left(ab+bc+ca\right)\le\left(a+b+c\right)^2\)
\(\Leftrightarrow\dfrac{1}{2}\left(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right)\ge0\) (đúng)
\(\Rightarrow ab+bc+ca\le\dfrac{\left(a+b+c\right)^2}{3}=1\)
Khi đó áp dụng BĐT Cauchy-Schwarz ta có:
\(\dfrac{a}{\sqrt{a^2+1}}\le\dfrac{a}{\sqrt{a^2+ab+bc+ca}}=\dfrac{a}{\sqrt{\left(a+b\right)\left(a+c\right)}}\)
\(\le\dfrac{1}{2}\left(\dfrac{a}{a+b}+\dfrac{a}{a+c}\right)\). Tương tự cho 2 BĐT còn lại:
\(\dfrac{b}{\sqrt{b^2+1}}\le\dfrac{1}{2}\left(\dfrac{b}{a+b}+\dfrac{b}{b+c}\right);\dfrac{c}{\sqrt{c^2+1}}\le\dfrac{1}{2}\left(\dfrac{c}{a+c}+\dfrac{c}{b+c}\right)\)
Cộng theo vế 3 BĐT trên ta có:
\(VT\le\dfrac{1}{2}\left(\dfrac{a+b}{a+b}+\dfrac{b+c}{b+c}+\dfrac{c+a}{c+a}\right)=\dfrac{3}{2}=VP\)
Xảy ra khi \(a=b=c=\dfrac{\sqrt{3}}{3}\)
Áp dụng BĐT Bu-nhi-a ta có:
\(\sqrt{a^2+1}=\sqrt{a^2+\dfrac{1}{3}+\dfrac{1}{3}+\dfrac{1}{3}}=\dfrac{1}{2}\sqrt{4\left(a^2+\dfrac{1}{3}+\dfrac{1}{3}+\dfrac{1}{3}\right)}\)
\(\ge\dfrac{1}{2}\sqrt{\left(a+\dfrac{1}{\sqrt{3}}.3\right)^2}=\dfrac{1}{2}\sqrt{\left(a+\sqrt{3}\right)^2}=\dfrac{a+\sqrt{3}}{2}\left(a>0\right)\)
Tương tự ta cũng có: \(\dfrac{b}{\sqrt{b^2+1}}\le\dfrac{2b}{b+\sqrt{3}}\)
\(\dfrac{c}{\sqrt{c^2+1}}\le\dfrac{2c}{c+\sqrt{3}}\)
=> \(\dfrac{a}{\sqrt{a^2+1}}+\dfrac{b}{\sqrt{b^2+1}}+\dfrac{c}{\sqrt{c^2+1}}\)
\(\le2\left(\dfrac{a}{2a+b+c}+\dfrac{b}{2b+a+c}+\dfrac{c}{2c+a+b}\right)\) (1)
Áp dụng BĐT phụ: \(\dfrac{1}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\ge\dfrac{1}{x+y}\) ta có:
\(\dfrac{a}{2a+b+c}+\dfrac{b}{2b+a+c}+\dfrac{c}{2c+a+b}\)
\(=\dfrac{a}{\left(a+b\right)+\left(a+c\right)}+\dfrac{b}{\left(a+b\right)+\left(b+c\right)}+\dfrac{c}{\left(a+c\right)+\left(b+c\right)}\)
\(\le\dfrac{1}{4}\left(\dfrac{a}{a+b}+\dfrac{a}{a+c}+\dfrac{b}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{a+c}+\dfrac{c}{b+c}\right)\)
\(=\dfrac{1}{4}\left(\dfrac{a+c}{a+c}+\dfrac{b+a}{a+b}+\dfrac{c+b}{b+c}\right)=\dfrac{3}{4}\) (2)
Từ (1); (2)
=> \(\dfrac{a}{\sqrt{a^2+1}}+\dfrac{b}{\sqrt{b^2+1}}+\dfrac{c}{\sqrt{c^2+1}}\le2.\dfrac{3}{4}=\dfrac{3}{2}\left(đpcm\right)\)
Dấu = xảy ra <=> \(a=b=c=\dfrac{1}{\sqrt{3}}\)