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\(P=\dfrac{bc}{\dfrac{a^2bc}{c}+\dfrac{a^2bc}{b}}+\dfrac{ca}{\dfrac{b^2ac}{a}+\dfrac{b^2ac}{c}}+\dfrac{ab}{\dfrac{c^2ab}{b}+\dfrac{c^2ab}{a}}=\dfrac{\left(bc\right)^2}{a^2b^2c+a^2bc^2}+\dfrac{\left(ca\right)^2}{b^2a^2c+b^2ac^2}+\dfrac{\left(ab\right)^2}{c^2a^2b+c^2ab^2}=\dfrac{\left(bc\right)^2}{ab+ac}+\dfrac{\left(ca\right)^2}{ba+bc}+\dfrac{\left(ab\right)^2}{ca+cb}\ge\dfrac{\left(ab+bc+ca\right)^2}{2\left(ab+bc+ca\right)}=\dfrac{ab+bc+ca}{2}\ge\dfrac{3\sqrt[3]{\left(abc\right)^2}}{2}=\dfrac{3}{2}\)
Dấu "=" xảy ra <=> a = b = c = 1
Lời giải:
Ta có: \(ab+bc+ac=4abc\Rightarrow \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=4\)
Áp dụng BĐT Bunhiacopxky:
\(\left(\frac{1}{a^4}+\frac{1}{b^4}+\frac{1}{c^4}\right)(1+1+1)\geq \left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)^2\) (1)
\(\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)(1+1+1)\geq \left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2\) (2)
Từ (1)(2) suy ra \(\frac{1}{a^4}+\frac{1}{b^4}+\frac{1}{c^4}\geq \frac{1}{27}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^4=\frac{4^4}{27}=\frac{256}{27}\)
Vậy \(P_{\min}=\frac{256}{27}\)
Dấu bằng xảy ra khi \(a=b=c=\frac{3}{4}\)
Ta có:
1+a2 = ab+bc+ca+a2 = a(a+b)+c(a+b)=(a+b)(a+c)
Tương tự: 1+b2 = (b+c)(b+a)
1+c2 = (c+a)(c+b)
\(\Rightarrow\) P = \(2a\sqrt{\dfrac{1}{\left(a+b\right)\left(a+c\right)}}+2b\sqrt{\dfrac{1}{\left(b+c\right)\left(b+a\right)}}+2c\sqrt{\dfrac{1}{\left(c+a\right)\left(c+b\right)}}\)
Áp dụng BĐT Cô-si ta có:
P\(\le\)\(a\left(\dfrac{1}{a+b}+\dfrac{1}{a+c}\right)+b\left(\dfrac{1}{4\left(b+c\right)}+\dfrac{1}{b+a}\right)+c\left(\dfrac{1}{4\left(c+b\right)}+\dfrac{1}{c+a}\right)\)\(\le\)\(\dfrac{a}{a+b}+\dfrac{a}{a+c}+\dfrac{b}{4\left(b+c\right)}+\dfrac{b}{b+a}+\dfrac{c}{4\left(c+b\right)}+\dfrac{c}{c+a}\)
= \(\dfrac{1}{4}+2=\dfrac{9}{4}\)
\(\Rightarrow\)Pmin = \(\dfrac{9}{4}\)
Dấu "=" xảy ra\(\Leftrightarrow\) b=c=\(\dfrac{a}{7}\)=\(\dfrac{\sqrt{15}}{15}\) \(\Rightarrow\) a = \(\dfrac{7\sqrt{15}}{15}\)
Lời giải:
Áp dụng BĐT AM-GM ngược dấu ta có:
\(A=\frac{ab}{\sqrt{c+ab}}+\frac{bc}{\sqrt{a+bc}}+\frac{ca}{\sqrt{b+ac}}=\frac{ab}{\sqrt{c(a+b+c)+ab}}+\frac{bc}{\sqrt{a(a+b+c)+bc}}+\frac{ca}{\sqrt{b(a+b+c)+ac}}\)
\(=\frac{ab}{\sqrt{(c+a)(c+b)}}+\frac{bc}{\sqrt{(a+b)(a+c)}}+\frac{ca}{\sqrt{(b+a)(b+c)}}\)
\(\leq \frac{1}{2}\left(\frac{ab}{c+a}+\frac{ab}{c+b}\right)+\frac{1}{2}\left(\frac{bc}{a+b}+\frac{bc}{a+c}\right)+\frac{1}{2}\left(\frac{ca}{b+a}+\frac{ca}{b+c}\right)\)
\(A\leq \frac{1}{2}\left(\frac{ab+bc}{a+c}+\frac{ab+ac}{b+c}+\frac{bc+ac}{a+b}\right)=\frac{1}{2}(b+a+c)=\frac{1}{2}\)
Vậy \(A_{\max}=\frac{1}{2}\) tại \(a=b=c=\frac{1}{3}\)
\(\sqrt{\dfrac{a+b}{c+ab}}+\sqrt{\dfrac{b+c}{a+bc}}+\sqrt{\dfrac{c+a}{b+ac}}\)
Bài này có xuất hiện rồi ,you vào mục tìm kiếm là thấy liền.
Lời giải vắn tắt:
\(A=\sum\sqrt{\dfrac{ab+2c^2}{1+ab-c^2}}=\sum\dfrac{ab+2c^2}{\sqrt{\left(ab+2c^2\right)\left(1+ab-c^2\right)}}\ge\sum\dfrac{2\left(ab+2c^2\right)}{1+2ab+c^2}=\sum\dfrac{2\left(ab+2c^2\right)}{\left(a+b\right)^2+2c^2}\ge\sum\dfrac{2\left(ab+2c^2\right)}{2\left(a^2+b^2\right)+2c^2}=\sum\left(ab+2c^2\right)=ab+bc+ca+2\)
( thay \(a^2+b^2+c^2=1\))
Đề sai rồi: a,b,c > 0 thì làm sao mà có: ab + bc + ca = 0 được.
\(P=\dfrac{bc}{a\left(b+c\right)}+\dfrac{ca}{b\left(c+a\right)}+\dfrac{ab}{c\left(a+b\right)}\)
\(=\dfrac{b^2c^2}{abc\left(b+c\right)}+\dfrac{c^2a^2}{abc\left(c+a\right)}+\dfrac{a^2b^2}{abc\left(a+b\right)}\)
\(\ge\dfrac{\left(ab+bc+ca\right)^2}{2abc\left(a+b+c\right)}\ge\dfrac{3abc\left(a+b+c\right)}{2abc\left(a+b+c\right)}=\dfrac{3}{2}\)
Dấu = xảy ra khi \(a=b=c\)
Có BĐT: \(a^2+b^2+c^2\ge ab+bc+ca\)
\(\Leftrightarrow\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\)
Ta có:
\(VT=\)\(\dfrac{1}{a^2+b^2+1}+\dfrac{1}{b^2+c^2+1}+\dfrac{1}{c^2+a^2+1}\)
\(=\dfrac{1+1+c^2}{\left(a^2+b^2+1\right)\left(1+1+c^2\right)}+\dfrac{1+1+a^2}{\left(b^2+c^2+1\right)\left(1+1+a^2\right)}+\dfrac{1+1+b^2}{\left(c^2+a^2+1\right)\left(1+1+b^2\right)}\)
Áp dụng BĐT Bunhiacopski cho mẫu số, ta có:
\(\left(a^2+b^2+c^2\right)\left(1+1+c^2\right)\ge\left(a+b+c\right)^2\)
\(\left(b^2+c^2+1\right)\left(1+1+a^2\right)\ge\left(b+c+a\right)^2\)
\(\left(c^2+a^2+1\right)\left(1+1+b^2\right)\ge\left(c+a+b\right)^2\)
\(\Rightarrow VT\le\dfrac{1+1+c^2}{\left(a+b+c\right)^2}+\dfrac{1+1+a^2}{\left(b+c+a\right)^2}+\dfrac{1+1+b^2}{\left(c+a+b\right)^2}=\dfrac{6+a^2+b^2+c^2}{\left(a+b+c\right)^2}\le\dfrac{6+ab+bc+ca}{3\left(ab+bc+ca\right)}=\dfrac{6+3}{3.3}=1\)
\("="\Leftrightarrow a=b=c=1\)