cho a,b>0 va a+b=1
cmr: (a+1/a)2+(b+1/b)2>=12,5
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\(ab+bc+ca\le1\)
\(\Rightarrow\sqrt{a^2+1}\ge\sqrt{a^2+ab+bc+ca}=\sqrt{\left(a+b\right)\left(a+c\right)}\)
\(\Rightarrow\dfrac{a}{\sqrt{a^2+1}}\le\dfrac{a}{\sqrt{\left(a+b\right)\left(a+c\right)}}\le\dfrac{\dfrac{a}{a+b}+\dfrac{a}{a+c}}{2}\)
\(tương\) \(tự\Rightarrow\Sigma\dfrac{a}{\sqrt{a^2+1}}\le\dfrac{\dfrac{a}{a+b}+\dfrac{a}{a+c}}{2}+\dfrac{\dfrac{b}{a+b}+\dfrac{b}{b+c}}{2}+\dfrac{\dfrac{c}{b+c}+\dfrac{c}{a+c}}{2}=\dfrac{3}{2}\left(đpcm\right)\)
\(dấu"="\Leftrightarrow a=b=c=\sqrt{\dfrac{1}{3}}\)
\(VT=\dfrac{a^4}{ab+ac}+\dfrac{b^4}{ab+bc}+\dfrac{c^4}{ac+bc}\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{2\left(ab+bc+ca\right)}\)
\(VT\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{2\left(a^2+b^2+c^2\right)}=\dfrac{1}{2}\)
Dấu "=" xảy ra khi \(a=b=c\)
Lời giải:
Áp dụng BĐT Bunhiacopxky:
\([(a+\frac{1}{a})^2+(b+\frac{1}{b})^2](1^2+1^2)\geq (a+\frac{1}{a}+b+\frac{1}{b})^2=(1+\frac{1}{a}+\frac{1}{b})^2\)
\(\Rightarrow (a+\frac{1}{a})^2+(b+\frac{1}{b})^2\geq \frac{1}{2}(1+\frac{1}{a}+\frac{1}{b})^2\)
Tiếp tục áp dụng BDDT Bunhiacopxky:
$\frac{1}{a}+\frac{1}{b}\geq \frac{4}{a+b}=4$
\(\Rightarrow (a+\frac{1}{a})^2+(b+\frac{1}{b})^2\geq \frac{1}{2}(1+\frac{1}{a}+\frac{1}{b})^2\geq \frac{1}{2}(1+4)^2=12,5\)
Ta có đpcm
Dấu "=" xảy ra khi $a=b=\frac{1}{2}$
\(a+b+c=1=>\left\{{}\begin{matrix}1-a=b+c\\1-b=a+c\\1-c=a+b\\\end{matrix}\right.\)
\(=>A=\left(\dfrac{1}{a}-1\right)\left(\dfrac{1}{b}-1\right)\left(\dfrac{1}{c}-1\right)=\left(\dfrac{1-a}{a}\right)\left(\dfrac{1-b}{b}\right)\left(\dfrac{1-c}{c}\right)\)
\(=\left(\dfrac{b+c}{a}\right)\left(\dfrac{a+c}{b}\right)\left(\dfrac{a+b}{c}\right)\)
bbđt AM-GM
\(=>A\ge\dfrac{2\sqrt{bc}.2\sqrt{ac}.2\sqrt{ab}}{abc}=\dfrac{8abc}{abc}=8\left(đpcm\right)\)
dấu"=" xảy ra<=>\(a=b=c=\dfrac{1}{3}\)
Đặt vế trái BĐT cần chứng minh là P
Ta có:
\(P=\left(\dfrac{a+b+c}{a}-1\right)\left(\dfrac{a+b+c}{b}-1\right)\left(\dfrac{a+b+c}{c}-1\right)\)
\(P=\dfrac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}\ge\dfrac{2\sqrt{ab}.2\sqrt{bc}.2\sqrt{ca}}{abc}=8\) (đpcm)
Dấu "=" xảy ra khi \(a=b=c=\dfrac{1}{3}\)
\(VT=\dfrac{1}{a+1}+\dfrac{1}{b+1}+\dfrac{1}{c+1}+\dfrac{2}{\left(a+1\right)^2}+\dfrac{2}{\left(b+1\right)^2}+\dfrac{2}{\left(c+1\right)^2}\)
Mặt khác:
\(\dfrac{1}{\left(\sqrt{ab}.\sqrt{\dfrac{a}{b}}+1.1\right)^2}+\dfrac{1}{\left(\sqrt{ab}.\sqrt{\dfrac{b}{a}}+1.1\right)^2}\ge\dfrac{1}{\left(1+ab\right)\left(1+\dfrac{a}{b}\right)}+\dfrac{1}{\left(1+ab\right)\left(1+\dfrac{b}{a}\right)}=\dfrac{1}{1+ab}\)
Do đó:
\(VT\ge\dfrac{1}{a+1}+\dfrac{1}{b+1}+\dfrac{1}{c+1}+\dfrac{1}{1+ab}+\dfrac{1}{1+bc}+\dfrac{1}{1+ca}\)
\(VT\ge\dfrac{1}{a+1}+\dfrac{1}{b+1}+\dfrac{1}{c+1}+\dfrac{1}{1+\dfrac{1}{c}}+\dfrac{1}{1+\dfrac{1}{a}}+\dfrac{1}{1+\dfrac{1}{b}}=3\)
Dấu "=" xảy ra khi \(a=b=c=1\)
cho em hỏi một tí ạ
Chộ \(\dfrac{1}{\left(\sqrt{ab}.\sqrt{\dfrac{a}{b}}+1.1\right)^2}+\dfrac{1}{\left(\sqrt{ab}.\sqrt{\dfrac{b}{a}}+1.1\right)^2}\ge\dfrac{1}{\left(ab+1\right)\left(1+\dfrac{a}{b}\right)}+\dfrac{1}{\left(1+ab\right)\left(1+\dfrac{b}{a}\right)}\)
áp dụng công thức gì đây ạ
BĐT tương đương
\(\dfrac{a+c}{b+c}-\dfrac{a}{b}>0\Leftrightarrow\dfrac{ab+bc-ab-ac}{b\left(b+c\right)}>0\)
\(\Leftrightarrow\dfrac{c\left(b-a\right)}{b\left(b+c\right)}>0\)\(\Leftrightarrow b-a>0\Leftrightarrow b>a\Leftrightarrow\dfrac{a}{b}< 1\)(đúng vì GT)
Đề sai rồi bạn ơi!