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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:
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 \(a=b=c=\frac{1}{\sqrt{3}}\)
Ko lq nhưng ta chuẩn hóa \(a+b+c=3\). So:
\(M\ge\dfrac{\left(a+b+c\right)^2}{2\left(a+b+c\right)}=\dfrac{3}{2}\)
Ta có:
\(\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2=9\\ \Leftrightarrow a+b+c+2\sqrt{ab}+2\sqrt{bc}+2\sqrt{ac}=9\\ \Leftrightarrow\sqrt{ab}+\sqrt{bc}+\sqrt{ac}=2\)
\(\Rightarrow\dfrac{\sqrt{a}}{a+2}+\dfrac{\sqrt{b}}{b+2}+\dfrac{\sqrt{c}}{c+2}=\dfrac{\sqrt{a}}{a+\sqrt{ab}+\sqrt{bc}+\sqrt{ac}}+\dfrac{\sqrt{b}}{b+\sqrt{ab}+\sqrt{bc}+\sqrt{ac}}+\dfrac{\sqrt{c}}{c+\sqrt{ab}+\sqrt{bc}+\sqrt{ac}}\\ =\dfrac{\sqrt{a}}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}+\sqrt{c}\right)}+\dfrac{\sqrt{b}}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{b}+\sqrt{c}\right)}+\dfrac{\sqrt{c}}{\left(\sqrt{b}+\sqrt{c}\right)\left(\sqrt{a}+\sqrt{c}\right)}\\ =\dfrac{\sqrt{a}\left(\sqrt{b}+\sqrt{c}\right)+\sqrt{b}\left(\sqrt{a}+\sqrt{c}\right)+\sqrt{c}\left(\sqrt{a}+\sqrt{b}\right)}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{b}+\sqrt{c}\right)\left(\sqrt{a}+\sqrt{c}\right)}\\ =\dfrac{2\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ac}\right)}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{b}+\sqrt{c}\right)\left(\sqrt{a}+\sqrt{c}\right)}\\ =\dfrac{4}{\sqrt{\left(\sqrt{a}+\sqrt{b}\right)^2\left(\sqrt{b}+\sqrt{c}\right)^2\left(\sqrt{a}+\sqrt{c}\right)^2}}\)\(=\dfrac{4}{\sqrt{\left(a+\sqrt{ab}+\sqrt{bc}+\sqrt{ac}\right)\left(b+\sqrt{ab}+\sqrt{bc}+\sqrt{ac}\right)\left(c+\sqrt{ab}+\sqrt{bc}+\sqrt{ac}\right)}}\\ =\dfrac{4}{\sqrt{\left(a+2\right)\left(b+2\right)\left(c+2\right)}}\)
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:
\(\dfrac{1}{2a+b+c}+\dfrac{1}{a+2b+c}+\dfrac{1}{a+b+2c}\le\dfrac{1}{16}.\left(\dfrac{4}{a}+\dfrac{4}{b}+\dfrac{4}{c}\right)=1\left(đpcm\right)\)
Áp dụng BĐT phụ:
\(3\left(a^2+a^2+b^2\right)\ge\left(2a+b\right)^2\)
P=\(\sum\dfrac{a}{\sqrt{2a^2+b^2}+\sqrt{3}}\)
\(\Rightarrow\)\(\dfrac{1}{\sqrt{3}}P=\sum\dfrac{a}{\sqrt{3\left(a^2+a^2+b^2\right)}+3}\)
\(\Rightarrow\)\(\dfrac{1}{\sqrt{3}}P\le\sum\dfrac{a}{\sqrt{\left(2a+b\right)^2}+a+b+c}=\sum\dfrac{a}{3a+2b+c}\)
Xét M=\(\sum\dfrac{a}{3a+2b+c}\)
\(3-3M=\sum\dfrac{2b+c}{3a+2b+c}\)
\(\Rightarrow\)\(3-3M=\sum\dfrac{\left(2b+c\right)^2}{\left(3a+2b+c\right)\left(2b+c\right)}\ge\)\(\dfrac{\left(3a+3b+3c\right)^2}{\sum\left(3a+2b+c\right)\left(2b+c\right)}\)
Mà
\(\sum\left(3a+2b+c\right)\left(2b+c\right)=5a^2+5b^2+5c^2+13ab+13bc+13ac=5\left(a+b+c\right)^2+3\left(ab+bc+ac\right)\le5\left(a+b+c\right)^2+\left(a+b+c\right)^2\)
\(\Rightarrow\)\(3-3M\ge\dfrac{\left(3a+3b+3c\right)^2}{6\left(a+b+c\right)^2}\ge\dfrac{9}{6}=\dfrac{3}{2}\)
\(\Rightarrow\)\(M\le\dfrac{1}{2}\)
\(\Rightarrow\)\(\dfrac{1}{\sqrt{3}}P\le\dfrac{1}{2}\Rightarrow P\le\dfrac{\sqrt{3}}{2}\)
Lời giải:
\(a+b+c=abc\Rightarrow a(a+b+c)=a^2bc\)
\(\Rightarrow a(a+b+c)+bc=bc(a^2+1)\)
\(\Leftrightarrow (a+b)(a+c)=bc(a^2+1)\)
\(\Leftrightarrow a^2+1=\frac{(a+b)(a+c)}{bc}\Rightarrow \frac{1}{\sqrt{a^2+1}}=\sqrt{\frac{bc}{(a+b)(a+c)}}\)
Áp dụng BĐT AM-GM:
\(\frac{1}{\sqrt{a^2+1}}=\sqrt{\frac{bc}{(a+b)(a+c)}}\leq \frac{1}{2}(\frac{b}{a+b}+\frac{c}{a+c})\)
Hoàn toàn tương tự:
\(\frac{1}{\sqrt{b^2+1}}=\sqrt{\frac{ac}{(b+a)(b+c)}}\leq \frac{1}{2}(\frac{a}{b+a}+\frac{c}{b+c})\)
\(\frac{1}{\sqrt{c^2+1}}=\sqrt{\frac{ab}{(c+a)(c+b)}}\leq \frac{1}{2}(\frac{a}{c+a}+\frac{b}{b+c})\)
Cộng theo vế:
\(\Rightarrow \frac{1}{\sqrt{a^2+1}}+\frac{1}{\sqrt{b^2+1}}+\frac{1}{\sqrt{c^2+1}}\leq \frac{1}{2}(\frac{a+b}{a+b}+\frac{b+c}{b+c}+\frac{c+a}{c+a})=\frac{3}{2}\)
Ta có đpcm
Dấu "=" xảy ra khi $a=b=c=\sqrt{3}$
Áp dụng BĐT Cauchy-Schwarz ta có:
\(\left(a^2+b+c\right)\left(1+b+c\right)\ge\left(a+b+c\right)^2\)
\(\Rightarrow a^2+b+c\ge\dfrac{\left(a+b+c\right)^2}{1+b+c}\Rightarrow\sqrt{\dfrac{a^2}{a^2+b+c}}\le\dfrac{a\sqrt{1+b+c}}{a+b+c}\)
Tương tự cho 2 BĐT còn lại rồi cộng theo vế:
\(A\le\dfrac{a\sqrt{1+b+c}+b\sqrt{1+c+a}+c\sqrt{1+a+b}}{a+b+c}\)
Áp dụng BĐT AM-GM ta có:
\(a\sqrt{1+b+c}=\dfrac{\sqrt{3a}\sqrt{a+ab+bc}}{\sqrt{3}}\le\dfrac{4a+ab+bc}{2\sqrt{3}}\)
Tương tự cho 2 BĐT còn lại rồi cộng theo vế:
\(\Rightarrow\dfrac{a\sqrt{1+b+c}+b\sqrt{1+c+a}+c\sqrt{1+a+b}}{a+b+c}\le\dfrac{2(a+b+c)+(ab+bc+ca)}{\sqrt{3}(a+b+c)}\)
\(\le\dfrac{2(a+b+c)+\dfrac{(a+b+c)^2}{3}}{\sqrt{3}(a+b+c)}\le\dfrac{2+\dfrac{a+b+c}{3}}{\sqrt{3}}\le\sqrt{3}\)
Hay \(A\le\sqrt{3}\) *ĐPCM*
đề sai