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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:
\(\dfrac{1}{2a+b+c}+\dfrac{1}{a+2b+c}+\dfrac{1}{a+b+2c}\le\dfrac{1}{16}.\left(\dfrac{4}{a}+\dfrac{4}{b}+\dfrac{4}{c}\right)=1\left(đpcm\right)\)
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:
\(\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2=9\\ \Leftrightarrow a+b+c+2\sqrt{ab}+2\sqrt{bc}+2\sqrt{ac}=9\\ \Leftrightarrow\sqrt{ab}+\sqrt{bc}+\sqrt{ac}=2\)
\(\Rightarrow\dfrac{\sqrt{a}}{a+2}+\dfrac{\sqrt{b}}{b+2}+\dfrac{\sqrt{c}}{c+2}=\dfrac{\sqrt{a}}{a+\sqrt{ab}+\sqrt{bc}+\sqrt{ac}}+\dfrac{\sqrt{b}}{b+\sqrt{ab}+\sqrt{bc}+\sqrt{ac}}+\dfrac{\sqrt{c}}{c+\sqrt{ab}+\sqrt{bc}+\sqrt{ac}}\\ =\dfrac{\sqrt{a}}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}+\sqrt{c}\right)}+\dfrac{\sqrt{b}}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{b}+\sqrt{c}\right)}+\dfrac{\sqrt{c}}{\left(\sqrt{b}+\sqrt{c}\right)\left(\sqrt{a}+\sqrt{c}\right)}\\ =\dfrac{\sqrt{a}\left(\sqrt{b}+\sqrt{c}\right)+\sqrt{b}\left(\sqrt{a}+\sqrt{c}\right)+\sqrt{c}\left(\sqrt{a}+\sqrt{b}\right)}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{b}+\sqrt{c}\right)\left(\sqrt{a}+\sqrt{c}\right)}\\ =\dfrac{2\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ac}\right)}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{b}+\sqrt{c}\right)\left(\sqrt{a}+\sqrt{c}\right)}\\ =\dfrac{4}{\sqrt{\left(\sqrt{a}+\sqrt{b}\right)^2\left(\sqrt{b}+\sqrt{c}\right)^2\left(\sqrt{a}+\sqrt{c}\right)^2}}\)\(=\dfrac{4}{\sqrt{\left(a+\sqrt{ab}+\sqrt{bc}+\sqrt{ac}\right)\left(b+\sqrt{ab}+\sqrt{bc}+\sqrt{ac}\right)\left(c+\sqrt{ab}+\sqrt{bc}+\sqrt{ac}\right)}}\\ =\dfrac{4}{\sqrt{\left(a+2\right)\left(b+2\right)\left(c+2\right)}}\)
b)CM: \(ab\sqrt{1+\dfrac{1}{a^2b^2}}-\sqrt{a^2b^2+1}=0\)
\(VT=ab\sqrt{\dfrac{a^2b^2+1}{\left(ab\right)^2}}-\sqrt{a^2b^2+1}\)
\(VT=ab\dfrac{\sqrt{a^2b^2+1}}{ab}-\sqrt{a^2b^2+1}\)
\(VT=\sqrt{a^2b^2+1}-\sqrt{a^2b^2+1}\)
\(VT=0=VP\)
nhầm mọi người ơi chứng minh cho mình <=\(\dfrac{3}{\sqrt{2}}\)
Áp dụng \(x^2+y^2+z^2\ge xy+yz+zx\) và \(x^2+y^2+z^2\ge\dfrac{1}{3}\left(x+y+z\right)^2\)
\(N\ge\dfrac{a^2b}{c}+\dfrac{b^2c}{a}+\dfrac{c^2a}{b}\ge\dfrac{1}{3}\left(a\sqrt{\dfrac{b}{c}}+b\sqrt{\dfrac{c}{a}}+c\sqrt{\dfrac{a}{b}}\right)^2=3\) (đpcm)
Dấu "=" xảy ra khi \(a=b=c=1\)
thx, appreciate it