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Lời giải:
Do \(ab+bc+ac=1\) nên:
\(a^2+1=a^2+ab+bc+ac=(a+b)(a+c)\)
\(b^2+1=b^2+ab+bc+ac=(b+a)(b+c)\)
\(c^2+1=c^2+ab+bc+ac=(c+a)(c+b)\)
Do đó:
\(A=a\sqrt{\frac{(b^2+1)(c^2+1)}{a^2+1}}+b\sqrt{\frac{(a^2+1)(c^2+1)}{b^2+1}}+c\sqrt{\frac{(b^2+1)(a^2+1)}{c^2+1}}\)
\(=a\sqrt{\frac{(b+c)(b+a)(c+a)(c+b)}{(a+b)(a+c)}}+b\sqrt{\frac{(a+b)(a+c)(c+a)(c+b)}{(b+a)(b+c)}}+c\sqrt{\frac{(b+a)(b+c)(a+b)(a+c)}{(c+a)(c+b)}}\)
\(=a(b+c)+b(a+c)+c(a+b)=2(ab+bc+ac)=2\)
Vậy \(A=2\)
ta có :\(a^2-ab+b^2=\left(a+b\right)^2-3ab\ge\left(a+b\right)^2-\dfrac{3}{4}\left(a+b\right)^2=\dfrac{1}{4}\left(a+b\right)^2\)(theo BĐT AM-GM)
\(\Rightarrow P\ge\sum\dfrac{a+b}{2\sqrt{ab+1}}\)
ÁP dụng BĐT AM-GM:
\(\dfrac{a+b}{2\sqrt{ab+1}}+\dfrac{b+c}{2\sqrt{bc+1}}+\dfrac{c+a}{2\sqrt{ca+1}}\ge3\sqrt[3]{\dfrac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{8\sqrt{\left(ab+1\right)\left(bc+1\right)\left(ca+1\right)}}}=\dfrac{3}{2}.\dfrac{1}{\sqrt[3]{\sqrt{\left(ab+1\right)\left(bc+1\right)\left(ca+1\right)}}}\)
Mà \(\sqrt[3]{\left(ab+1\right)\left(bc+1\right)\left(ca+1\right)}\le\dfrac{1}{3}\left(ab+bc+ca+3\right)\)
\(\Rightarrow P\ge\dfrac{3\sqrt{3}}{2\sqrt{\left(ab+bc+ca+3\right)}}\)(*)
ta liên tưởng đến BĐT phụ:\(\left(x+y\right)\left(y+z\right)\left(z+x\right)\ge\dfrac{8}{9}\left(x+y+z\right)\left(xy+yz+xz\right)\)
Cm: phân tích :\(VT=xy\left(x+y\right)+yz\left(y+z\right)+zx\left(x+z\right)+2xyz\)
\(=xy\left(x+y\right)+yz\left(y+z\right)+xz\left(z+x\right)+3xyz-xyz\)
\(=\left(x+y+z\right)\left(xy+yz+xz\right)-xyz\)
mà \(\left(x+y+z\right)\left(xy+yz+xz\right)\ge3\sqrt[3]{xyz}.3\sqrt[3]{x^2y^2z^2}=9xyz\)
nên \(\left(x+y\right)\left(y+z\right)\left(z+x\right)\ge\left(x+y+z\right)\left(xy+yz+xz\right)-\dfrac{1}{9}\left(x+y+z\right)\left(xy+yz+xz\right)=\dfrac{8}{9}\left(x+y+z\right)\left(xy+yz+zx\right)\)
Áp dụng:
\(1=\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge\dfrac{8}{9}\left(a+b+c\right)\left(ab+bc+ca\right)\)
mặt khác,theo AM-GM,dễ dàng chứng minh được \(a+b+c\ge\dfrac{3}{2}\)
nên \(1\ge\dfrac{8}{9}.\dfrac{3}{2}\left(ab+bc+ca\right)\Leftrightarrow ab+bc+ca\le\dfrac{3}{4}\)
từ (*)\(\Rightarrow P\ge\dfrac{3\sqrt{3}}{2\sqrt{\dfrac{3}{4}+3}}=\dfrac{3}{\sqrt{5}}\)
Dấu = xảy ra khi \(a=b=c=\dfrac{1}{2}\)
Xét \(\sqrt{\dfrac{\left(a+bc\right)\left(b+ac\right)}{c+ab}}=\sqrt{\dfrac{\left(a\left(a+b+c\right)+bc\right)\left(b\left(a+b+c\right)+ac\right)}{c\left(a+b+c\right)+ab}}\)
\(=\sqrt{\dfrac{\left(a^2+ab+ac+bc\right)\left(ab+b^2+bc+ac\right)}{ac+bc+c^2+ab}}\)
\(=\sqrt{\dfrac{\left(a+b\right)\left(a+c\right)\left(a+b\right)\left(b+c\right)}{\left(a+c\right)\left(b+c\right)}}\)\(=\sqrt{\left(a+b\right)^2}=a+b\)
Tương tự cho 2 đẳng thức còn lại rồi cộng theo vế
\(P=a+b+b+c+c+a=2\left(a+b+c\right)=2\)
Lời giải:
\(a+b+c=abc\)
\(\Rightarrow a(a+b+c)=a^2bc\)
\(\Rightarrow a(a+b+c)+bc=a^2bc+bc\)
\(\Rightarrow (a+b)(a+c)=bc(a^2+1)\)
\(\Rightarrow \frac{a}{\sqrt{bc(a^2+1)}}=\frac{a}{\sqrt{(a+b)(a+c)}}\leq \frac{1}{2}\left(\frac{a}{a+b}+\frac{a}{a+c}\right)\) (theo BĐT AM-GM ngược dấu)
Hoàn toàn tương tự:
\(\frac{b}{\sqrt{ca(b^2+1)}}\leq \frac{1}{2}\left(\frac{b}{b+a}+\frac{b}{b+c}\right)\)
\(\frac{c}{\sqrt{ab(c^2+1)}}\leq \frac{1}{2}\left(\frac{c}{c+a}+\frac{c}{c+b}\right)\)
Cộng theo vế những BĐT thu được ở trên ta có:
\(S\leq \frac{1}{2}\left(\frac{a+b}{a+b}+\frac{b+c}{b+c}+\frac{c+a}{c+a}\right)=\frac{3}{2}\)
Vậy \(S_{\max}=\frac{3}{2}\Leftrightarrow a=b=c=\sqrt{3}\)
bai nay t lam roi vao trang chu cua nick thangbnsh cua t keo xuong tim la thay
Câu hỏi của Tuyển Trần Thị - Toán lớp 9 | Học trực tuyến
a) CM:\(\sqrt{\left(n+1\right)^2}+\sqrt{n^2}=\left(n+1\right)^2-n^2\)
\(\Leftrightarrow n+1+n=\left(n+1-n\right)\left(n+1+n\right)\)
\(\Leftrightarrow2n+1=1\left(2n+1\right)\)
\(\Leftrightarrow2n+1=2n+1\)
\(\Rightarrow\sqrt{\left(n+1\right)^2}+\sqrt{n^2}=\left(n+1\right)^2-n^2\)
Câu b) ý 2:
Áp dụng BĐT cô si ta có :
\(\dfrac{a}{b}+\dfrac{b}{c}\ge2\sqrt{\dfrac{a}{c}}\\ \dfrac{b}{c}+\dfrac{c}{a}\ge2\sqrt{\dfrac{b}{a}}\\ \dfrac{c}{a}+\dfrac{a}{b}\ge2\sqrt{\dfrac{c}{b}}\\ \Leftrightarrow2\left(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\right)\ge2\left(\sqrt{\dfrac{a}{c}}+\sqrt{\dfrac{b}{a}}+\sqrt{\dfrac{c}{b}}\right)\\ \Rightarrowđpcm\)
Đề sai rồi: a,b,c > 0 thì làm sao mà có: ab + bc + ca = 0 được.
mk viết nhầm
\(ab+bc+ca=1\)
bn giúp mk với