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Lời giải:
Theo hệ quả quen thuộc của BĐT AM-GM thì:
\((a+b+c)^2\geq 3(ab+bc+ac)\)
\(\Leftrightarrow (\sqrt{3})^2\geq 3(ab+bc+ac)\Rightarrow ab+bc+ac\leq 1\)
\(\Rightarrow \frac{a}{\sqrt{a^2+1}}\leq \frac{a}{\sqrt{a^2+ab+bc+ac}}=\frac{a}{\sqrt{(a+b)(a+c)}}\)
Hoàn toàn TT với các phân thức còn lại và cộng theo vế:
\(\Rightarrow \text{VT}\leq \frac{a}{\sqrt{(a+b)(a+c)}}+\frac{b}{\sqrt{(b+c)(b+a)}}+\frac{c}{\sqrt{(c+a)(c+b)}}\)
\(\leq \frac{1}{2}\left(\frac{a}{a+b}+\frac{a}{a+c}\right)+\frac{1}{2}\left(\frac{b}{b+c}+\frac{b}{b+a}\right)+\frac{1}{2}\left(\frac{c}{c+a}+\frac{c}{c+b}\right)\) (BĐT Cauchy)
hay \(\text{VT}\leq \frac{1}{2}\left(\frac{a+b}{a+b}+\frac{b+c}{b+c}+\frac{c+a}{c+a}\right)=\frac{3}{2}\)(đpcm)
Dấu "=" xảy ra khi \(a=b=c=\frac{1}{\sqrt{3}}\)
a) điều kiện xác định : \(a>2;a\ne11\)
ta có : \(P=\left(\dfrac{\sqrt{a-2}+2}{3}\right)\left(\dfrac{\sqrt{a-2}}{3+\sqrt{a-2}}+\dfrac{a+7}{11-a}\right):\left(\dfrac{3\sqrt{a-2}+1}{a-3\sqrt{a-2}-2}-\dfrac{1}{\sqrt{a-2}}\right)\)
\(\Leftrightarrow P=\left(\dfrac{\sqrt{a-2}+2}{3}\right)\left(\dfrac{\sqrt{a-2}}{3+\sqrt{a-2}}+\dfrac{a+7}{\left(3+\sqrt{a-2}\right)\left(3-\sqrt{a-2}\right)}\right):\left(\dfrac{3\sqrt{a-2}+1}{\sqrt{a-2}\left(\sqrt{a-2}-3\right)}-\dfrac{1}{\sqrt{a-2}}\right)\) \(\Leftrightarrow P=\left(\dfrac{\sqrt{a-2}+2}{3}\right)\left(\dfrac{\sqrt{a-2}\left(3-\sqrt{a-2}\right)+a+7}{\left(3+\sqrt{a-2}\right)\left(3-\sqrt{a-2}\right)}\right):\left(\dfrac{3\sqrt{a-2}+1-\sqrt{a-2}+3}{\sqrt{a-2}\left(\sqrt{a-2}-3\right)}\right)\) \(\Leftrightarrow P=\left(\dfrac{\sqrt{a-2}+2}{3}\right)\left(\dfrac{3\left(\sqrt{a-2}+3\right)}{\left(3+\sqrt{a-2}\right)\left(3-\sqrt{a-2}\right)}\right):\left(\dfrac{2\sqrt{a-2}+4}{\sqrt{a-2}\left(\sqrt{a-2}-3\right)}\right)\) \(\Leftrightarrow P=\left(\dfrac{\sqrt{a-2}+2}{3}\right)\left(\dfrac{3}{\left(3-\sqrt{a-2}\right)}\right)\left(\dfrac{\sqrt{a-2}\left(\sqrt{a-2}-3\right)}{2\left(\sqrt{a-2}+2\right)}\right)\) \(\Leftrightarrow P=\dfrac{-\sqrt{a-2}}{2}\)
ta có : \(a+b=\sqrt{2017-a^2}+\sqrt{2017-b^2}\)
\(\Leftrightarrow\left(a+b\right)\left(\sqrt{2017-a^2}-\sqrt{2017-b^2}\right)=b^2-a^2\)
\(\Leftrightarrow b-a=\sqrt{2017-a^2}-\sqrt{2017-b^2}\)
\(\Leftrightarrow2b=2\sqrt{2017-a^2}\Leftrightarrow b^2=2017-a^2\Rightarrow\left(đpcm\right)\)
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 <=> \(a=b=c=\dfrac{1}{\sqrt{3}}\)
Lời giải:
\(a+b+c=abc\Rightarrow a(a+b+c)=a^2bc\)
\(\Rightarrow a(a+b+c)+bc=bc(a^2+1)\)
\(\Leftrightarrow (a+b)(a+c)=bc(a^2+1)\Rightarrow a^2+1=\frac{(a+b)(a+c)}{bc}\)
\(\Rightarrow \frac{1}{\sqrt{a^2+1}}=\sqrt{\frac{bc}{(a+b)(a+c)}}\)
Hoàn toàn tương tự với các phân thức còn lại
\(\Rightarrow \text{VT}=\frac{1}{\sqrt{a^2+1}}+\frac{1}{\sqrt{b^2+1}}+\frac{1}{\sqrt{c^2+1}}=\sqrt{\frac{bc}{(a+b)(a+c)}}+\sqrt{\frac{ac}{(b+a)(b+c)}}+\sqrt{\frac{ab}{(c+a)(c+b)}}\)
Áp dụng BĐT Cauchy:
\(\sqrt{\frac{bc}{(a+b)(a+c)}}+\sqrt{\frac{ac}{(b+a)(b+c)}}+\sqrt{\frac{ab}{(c+a)(c+b)}}\leq \frac{1}{2}\left(\frac{b}{a+b}+\frac{c}{a+c}\right)+\frac{1}{2}\left(\frac{a}{b+a}+\frac{c}{b+c}\right)+\frac{1}{2}\left(\frac{a}{c+a}+\frac{b}{c+b}\right)\)
\(=\frac{1}{2}\left(\frac{b+a}{b+a}+\frac{c+b}{c+b}+\frac{c+a}{c+a}\right)=\frac{3}{2}\)
\(\Rightarrow \text{VT}\leq \frac{3}{2}\) (đpcm)
Dấu "=" xảy ra khi $a=b=c=\sqrt{3}$
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:
\(\dfrac{1}{2a+b+c}+\dfrac{1}{a+2b+c}+\dfrac{1}{a+b+2c}\le\dfrac{1}{16}.\left(\dfrac{4}{a}+\dfrac{4}{b}+\dfrac{4}{c}\right)=1\left(đpcm\right)\)
\(P=\dfrac{\left(5+2\sqrt{6}\right)\sqrt{5-2\sqrt{6}}}{\sqrt{3}+\sqrt{2}}\)
=\(\dfrac{\left(3+2\sqrt{2.3}+2\right)\sqrt{3-2\sqrt{3.2}+2}}{\sqrt{3}+\sqrt{2}}\)
=\(\dfrac{\left(\sqrt{3}+\sqrt{2}\right)^2\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}}{\sqrt{3}+\sqrt{2}}\)
=\(\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)\)
=\(3-2=1\)
ta có : \(\dfrac{1}{\sqrt{a}+\sqrt{b}}+\dfrac{1}{\sqrt{b}+\sqrt{c}}\ge2\sqrt{\dfrac{1}{\sqrt{ab}+\sqrt{ac}+\sqrt{bc}+b}}\)
\(\ge\dfrac{2}{\sqrt{a+b+c+b}}=\dfrac{2}{\sqrt{4b}}=\dfrac{2}{2\sqrt{b}}=\dfrac{1}{\sqrt{b}}=\dfrac{2}{\sqrt{a+c}}\ge\dfrac{2}{\sqrt{a}+\sqrt{b}}\)
dấu "=" xảy ra khi \(a=b=c\Leftrightarrow a+c=2b\Rightarrow\left(đpcm\right)\)
b) \(\dfrac{\sqrt{a}}{\sqrt{a}-\sqrt{b}}-\dfrac{\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\dfrac{2b}{a-b}\)
\(=\dfrac{\sqrt{a}}{\sqrt{a}-\sqrt{b}}-\dfrac{\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\dfrac{2b}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\dfrac{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)-\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)-2b}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\dfrac{a+\sqrt{ab}-\sqrt{ab}+b-\sqrt{ab}+b-2b}{a-b}\)
\(=\dfrac{a}{a-b}\)
bài này khó giúp hộ em với