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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:
\(\dfrac{1}{2a+b+c}+\dfrac{1}{a+2b+c}+\dfrac{1}{a+b+2c}\le\dfrac{1}{16}.\left(\dfrac{4}{a}+\dfrac{4}{b}+\dfrac{4}{c}\right)=1\left(đpcm\right)\)
Ta có: \(\dfrac{b}{\sqrt{a+b}-\sqrt{a-b}}=\dfrac{b}{\dfrac{\left(a+b\right)-\left(a-b\right)}{\sqrt{a+b}+\sqrt{a-b}}}\)
\(=\dfrac{b}{\dfrac{a+b-a+b}{\sqrt{a+b}+\sqrt{a-b}}}=\dfrac{\sqrt{a+b}+\sqrt{a-b}}{b}\)
Và \(\dfrac{c}{\sqrt{a+c}-\sqrt{a-c}}=\dfrac{c}{\dfrac{\left(a+c\right)-\left(a-c\right)}{\sqrt{a+c}+\sqrt{a-c}}}\)
\(=\dfrac{c}{\dfrac{a+c-a+c}{\sqrt{a+c}+\sqrt{a-c}}}=\dfrac{\sqrt{a+c}+\sqrt{a-c}}{c}\)
Từ \(a>b>c>0\) thì \(\left\{{}\begin{matrix}a+b>a+c\\a-b>a-c\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}\sqrt{a+b}>\sqrt{a+c}\\\sqrt{a-b}>\sqrt{a-c}\end{matrix}\right.\)
\(\Rightarrow\sqrt{a+b}+\sqrt{a-b}>\sqrt{a+c}+\sqrt{a-c}\)
\(\Rightarrow\dfrac{\sqrt{a+b}+\sqrt{a-b}}{b}< \dfrac{\sqrt{a+c}+\sqrt{a-c}}{c}\left(b>c>0\right)\)
Hay ta có ĐPCM
\(\dfrac{b}{\sqrt{a+b}-\sqrt{a-b}}=\dfrac{b}{\dfrac{a+b-a-b}{\sqrt{a+b}+\sqrt{a-b}}}=\dfrac{b}{\dfrac{0}{\sqrt{a+b}+\sqrt{a-b}}}\rightarrow\varnothing\)
Ta có \(\Sigma\sqrt{\dfrac{a}{b+c}}=\Sigma\dfrac{a}{\sqrt{a\left(b+c\right)}}\)
Theo AM-GM ta có
\(\Sigma\dfrac{a}{\sqrt{a\left(b+c\right)}}\ge\Sigma\dfrac{a}{\dfrac{a+b+c}{2}}=\Sigma\dfrac{2a}{a+b+c}=\dfrac{2\left(a+b+c\right)}{a+b+c}=2\)
dấu bằng xảy ra khi \(\left\{{}\begin{matrix}a=b+c\\b=a+c\\c=a+b\end{matrix}\right.\Rightarrow a+b+c=2\left(a+b+c\right)\Rightarrow1=2\) (vô lí)
nên\(\Sigma\sqrt{\dfrac{a}{b+c}}>2\)
Chắc đề bị nhầm rồi.
\(\dfrac{a}{\sqrt{b+1}}+\dfrac{b}{\sqrt{c+1}}+\dfrac{c}{\sqrt{a+1}}\ge2\sqrt{2}\left(\dfrac{a}{3+b}+\dfrac{b}{3+c}+\dfrac{c}{3+a}\right)\)
\(\ge2\sqrt{2}.\dfrac{\left(a+b+c\right)^2}{3\left(a+b+c\right)+\left(ab+bc+ca\right)}\ge2\sqrt{2}.\dfrac{9}{9+\dfrac{\left(a+b+c\right)^2}{3}}=2\sqrt{2}.\dfrac{9}{12}=\dfrac{3}{\sqrt{2}}\)
Lời giải:
Đặt \(\left ( \frac{\sqrt{a^2+b^2}}{c},\frac{\sqrt{b^2+c^2}}{a}, \frac{\sqrt{c^2+a^2}}{b} \right )=(x,y,z)\)
BĐT cần chứng minh tương đương với:
\(x+y+z\geq 2\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)\((*)\)
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Từ cách đặt $x,y,z$ ta có:
\(\frac{1}{x^2+1}+\frac{1}{y^2+1}+\frac{1}{z^2+1}=1\)
Áp dụng BĐT Bunhiacopxky:
\(\frac{x^2+1}{x^2}+\frac{y^2+1}{y^2}+\frac{z^2+1}{z^2}=\left(\frac{1}{x^2+1}+\frac{1}{y^2+1}+\frac{1}{z^2+1}\right)\left(\frac{x^2+1}{x^2}+\frac{y^2+1}{y^2}+\frac{z^2+1}{z^2}\right)\)
\(\geq \left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2\)
\(\Leftrightarrow 3\geq 2\left(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}\right)\)
\(\Leftrightarrow xyz\geq \frac{2}{3}(x+y+z)\)
\(\Rightarrow xyz(x+y+z)\geq \frac{2}{3}(x+y+z)^2\)
Áp dụng BĐT AM_GM ta lại có:
\((x+y+z)^2\geq 3(xy+yz+xz)\). Do đó:
\(xyz(x+y+z)\geq 2(xy+yz+xz)\)
\(\Leftrightarrow x+y+z\geq 2\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)
Đúng theo \((*)\)
Do đó ta có đpcm
Dấu bằng xảy ra khi \(a=b=c\)
áp dụng bat dang thuc bunhiacóki
ta có \(\dfrac{\sqrt{a^2+b^2}}{c}\ge\dfrac{a+b}{\sqrt{2}c}\)
ttu vt \(\ge\dfrac{1}{\sqrt{2}}\left(\dfrac{a+b}{c}+\dfrac{b+c}{a}+\dfrac{c+a}{b}\right)\)
=\(\dfrac{a}{\sqrt{2}}\left(\dfrac{1}{b}+\dfrac{1}{c}\right)+\dfrac{b}{\sqrt{2}}\left(\dfrac{1}{a}+\dfrac{1}{c}\right)+\dfrac{c}{\sqrt{2}}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\) (1)
áp dung bdt \(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\)
ta có (1) \(\ge\dfrac{a}{\sqrt{2}}.\dfrac{4}{b+c}\)
tiếp tục áp dụng bunhia ta có \(\dfrac{a}{\sqrt{2}}.\dfrac{4}{b+c}\ge\dfrac{a}{\sqrt{2}}.\dfrac{4}{\sqrt{2\left(b^2+c^2\right)}}=\dfrac{2a}{\sqrt{b^2+c^2}}\)
ttuong tu ta có \(vt\ge2\left(\dfrac{a}{\sqrt{b^2+c2}}+\dfrac{b}{\sqrt{a^2+c^2}}+\dfrac{c}{\sqrt{a^2+b^2}}\right)\left(dpcm\right)\)
_ Chứng minh VT <2 .
Với a,b,c > 0, ta có:
\(a< a+b\Rightarrow\dfrac{a}{a+b}< 1=\dfrac{c}{c}\Rightarrow\dfrac{a}{a+b}< \dfrac{a+c}{a+b+c}\) (1)
\(b< b+c\Rightarrow\dfrac{b}{b+c}< 1=\dfrac{a}{a}\Rightarrow\dfrac{b}{b+c}< \dfrac{a+b}{a+b+c}\) (2)
\(c< c+a\Rightarrow\dfrac{c}{c+a}< 1=\dfrac{b}{b}\Rightarrow\dfrac{c}{c+a}< \dfrac{c+b}{a+b+c}\) (3)
Từ (1) , (2) và (3), Cộng vế theo vế ta có:
\(\dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a}< \dfrac{2\left(a+b+c\right)}{a+b+c}=2\)(*)
_Chứng minh VP > 2.
Theo BĐT Cô-si, ta có:
\(\sqrt{\dfrac{b+c}{a}.1}\le\left(\dfrac{b+c}{a}+1\right):2=\dfrac{b+c+a}{2a}\)
Do vậy : \(\sqrt{\dfrac{a}{b+c}}\ge\dfrac{2a}{a+b+c}\)
Tương tự:\(\sqrt{\dfrac{b}{a+c}}\ge\dfrac{2b}{a+b+c},\sqrt{\dfrac{c}{a+b}}\ge\dfrac{2c}{a+b+c}\)
Cộng vế theo vế
\(\sqrt{\dfrac{a}{b+c}}+\sqrt{\dfrac{b}{a+c}}+\sqrt{\dfrac{c}{a+b}}\ge\dfrac{2\left(a+b+c\right)}{a+b+c}=2\)
Dấu ''='' xảy ra \(\left\{{}\begin{matrix}a=b+c\\b=a+c\\c=a+b\end{matrix}\right.\)
\(\Rightarrow a+b+c=0\) (trái với g/t a,b,c >0)
Vậy đẳng thức khong xảy ra dấu ''=''
\(\Rightarrow\sqrt{\dfrac{a}{b+c}}+\sqrt{\dfrac{b}{a+c}}+\sqrt{\dfrac{c}{a+b}}>2\) (**)
Từ (*) và (**) \(\Rightarrow\dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a}< \sqrt{\dfrac{a}{b+c}}+\sqrt{\dfrac{b}{a+c}}+\sqrt{\dfrac{c}{a+b}}\)
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
\(\sum\left(3a+2b+c\right)\left(2b+c\right)=5a^2+5b^2+5c^2+13ab+13bc+13ac=5\left(a+b+c\right)^2+3\left(ab+bc+ac\right)\le5\left(a+b+c\right)^2+\left(a+b+c\right)^2\)
\(\Rightarrow\)\(3-3M\ge\dfrac{\left(3a+3b+3c\right)^2}{6\left(a+b+c\right)^2}\ge\dfrac{9}{6}=\dfrac{3}{2}\)
\(\Rightarrow\)\(M\le\dfrac{1}{2}\)
\(\Rightarrow\)\(\dfrac{1}{\sqrt{3}}P\le\dfrac{1}{2}\Rightarrow P\le\dfrac{\sqrt{3}}{2}\)
Áp dụng bất đẳng thức Cô - si (a, b, c) dương ta có:
+/ \(\sqrt{\dfrac{b+c}{a}}=\sqrt{\dfrac{b+c}{a}.1}\le(\dfrac{b+c}{a}+1):2=\dfrac{a+b+c}{2a}\)
\(\Rightarrow\sqrt{\dfrac{a}{b+c}}\ge\dfrac{2a}{a+b+c}\)
+/ \(\sqrt{\dfrac{a+c}{b}}=\sqrt{\dfrac{a+c}{b}.1}\le(\dfrac{a+c}{b}+1):2=\dfrac{a+b+c}{2b}\)
\(\Rightarrow\sqrt{\dfrac{b}{a+c}}\ge\dfrac{2b}{a+b+c}\)+/ \(\sqrt{\dfrac{a+b}{c}}=\sqrt{\dfrac{a+b}{c}.1}\le(\dfrac{a+b}{c}+1):2=\dfrac{a+b+c}{2c}\)
\(\Rightarrow\sqrt{\dfrac{c}{a+b}}\ge\dfrac{2c}{a+b+c}\)
Khi đó:\(\sqrt{\dfrac{a}{b+c}}+\sqrt{\dfrac{b}{a+c}}+\sqrt{\dfrac{c}{a+b}}\ge\dfrac{2a+2b+2c}{a+b+c}=2\)
Dấu " = " xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}a=b+c\\b=a+c\\c=a+b\end{matrix}\right.\)
\(\Leftrightarrow a+b+c=0\) (Trái với giả thiết a, b, c là 3 số dương)
=> Đẳng thức không xảy ra
=> \(\sqrt{\dfrac{a}{b+c}}+\sqrt{\dfrac{b}{a+c}}+\sqrt{\dfrac{c}{a+b}}>2\) (đpcm)
Theo bất đẳng thức cô si, có:
\(\sqrt{1.\dfrac{b+c}{a}}\le\left(1+\dfrac{b+c}{a}\right):2=\dfrac{a+b+c}{2a}\)
\(\Rightarrow\sqrt{\dfrac{a}{b+c}}\ge\dfrac{2a}{a+b+c}~~~~~\left(1\right)\)
Tương tự: \(\sqrt{\dfrac{b}{c+a}}\ge\dfrac{2b}{a+b+c}~~~~~\left(2\right)\)
\(\sqrt{\dfrac{c}{a+b}}\ge\dfrac{2c}{a+b+c}~~~~~\left(3\right)\)
Cộng vế theo vế \(\left(1\right);\left(2\right);\left(3\right)\), ta có:
\(\sqrt{\dfrac{a}{b+c}}+\sqrt{\dfrac{b}{a+c}}+\sqrt{\dfrac{c}{a+b}}\ge\dfrac{2a}{a+b+c}+\dfrac{2b}{a+b+c}+\dfrac{2c}{a+b+c}=2\)