Cho 3 số thực a, b, c khác 0 thoả mãn ab+bc+ca=1. Tính giá trị của biểu thức: \(A=\dfrac{b^2+bc}{\sqrt{a^4+a^2}}.\dfrac{c^2+ca}{\sqrt{b^4+b^2}}.\dfrac{a^2+ab}{\sqrt{c^4+c^2}}\)
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NV
Nguyễn Việt Lâm
Giáo viên
13 tháng 6 2021
\(\sqrt{\dfrac{ab}{c+ab}}=\sqrt{\dfrac{ab}{c\left(a+b+c\right)+ab}}=\sqrt{\dfrac{ab}{\left(a+c\right)\left(b+c\right)}}\le\dfrac{1}{2}\left(\dfrac{a}{a+c}+\dfrac{b}{b+c}\right)\)
Tương tự: \(\sqrt{\dfrac{bc}{a+bc}}\le\dfrac{1}{2}\left(\dfrac{b}{a+b}+\dfrac{c}{a+c}\right)\) ; \(\sqrt{\dfrac{ca}{b+ca}}\le\dfrac{1}{2}\left(\dfrac{a}{a+b}+\dfrac{c}{b+c}\right)\)
Cộng vế với vế:
\(P\le\dfrac{1}{2}\left(\dfrac{a}{a+c}+\dfrac{c}{a+c}+\dfrac{b}{b+c}+\dfrac{c}{b+c}+\dfrac{b}{a+b}+\dfrac{a}{a+b}\right)=\dfrac{3}{2}\)
Dấu "=" xảy ra khi \(a=b=c=\dfrac{1}{3}\)
1 tháng 9 2021
Cho a, b, c, d là các chữ số thỏa mãn: ab+ca=da ab-ca=a Tìm giá trị của d.
30 tháng 8 2021
\(VT=\sqrt{\dfrac{a^2b^2}{c\left(a+b+c\right)+ab}}+\sqrt{\dfrac{b^2c^2}{a\left(a+b+c\right)+bc}}+\sqrt{\dfrac{a^2c^2}{b\left(a+b+c\right)+ac}}\\ VT=\sqrt{\dfrac{a^2b^2}{ac+ab+bc+c^2}}+\sqrt{\dfrac{b^2c^2}{a^2+ac+ab+bc}}+\sqrt{\dfrac{a^2c^2}{ab+bc+b^2+ac}}\\ VT=\sqrt{\dfrac{a^2b^2}{\left(c+a\right)\left(b+c\right)}}+\sqrt{\dfrac{a^2c^2}{\left(b+c\right)\left(a+b\right)}}+\sqrt{\dfrac{b^2c^2}{\left(a+b\right)\left(a+c\right)}}\)
Áp dụng BĐT Cauchy-Schwarz:
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{\dfrac{b^2c^2}{\left(a+b\right)\left(a+c\right)}}\le\dfrac{\dfrac{bc}{a+b}+\dfrac{bc}{a+c}}{2}\\\sqrt{\dfrac{a^2c^2}{\left(a+b\right)\left(b+c\right)}}\le\dfrac{\dfrac{ca}{a+b}+\dfrac{ca}{b+c}}{2}\\\sqrt{\dfrac{a^2b^2}{\left(b+c\right)\left(a+c\right)}}\le\dfrac{\dfrac{ab}{b+c}+\dfrac{ab}{a+c}}{2}\end{matrix}\right.\)
\(\Rightarrow VT\le\dfrac{\left(\dfrac{bc}{a+b}+\dfrac{ca}{a+b}\right)+\left(\dfrac{ca}{b+c}+\dfrac{ab}{b+c}\right)+\left(\dfrac{bc}{a+c}+\dfrac{ab}{a+c}\right)}{2}\\ \Rightarrow VT\le\dfrac{a+b+c}{2}=\dfrac{2}{2}=1\)
Dấu \("="\Leftrightarrow a=b=c=\dfrac{2}{3}\)
NV
Nguyễn Việt Lâm
Giáo viên
13 tháng 11 2021
\(\dfrac{ab}{a+b}=\dfrac{bc}{b+c}=\dfrac{ca}{c+a}\Rightarrow\dfrac{a+b}{ab}=\dfrac{b+c}{bc}=\dfrac{c+a}{ca}\)
\(\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{c}+\dfrac{1}{a}\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{a}=\dfrac{1}{c}\\\dfrac{1}{a}=\dfrac{1}{b}\end{matrix}\right.\) \(\Rightarrow a=b=c\)
\(\Rightarrow M=\dfrac{a^2+a^2+a^2}{a^2+a^2+a^2}=1\)
5 tháng 1 2023
- Theo BĐT Cauchy ta có:
\(\sqrt{a.1}\le\dfrac{a+1}{2}\)
\(\sqrt{b.1}\le\dfrac{b+1}{2}\)
\(\sqrt{c.1}\le\dfrac{c+1}{2}\)
\(\sqrt{ab}\le\dfrac{a+b}{2}\)
\(\sqrt{bc}\le\dfrac{b+c}{2}\)
\(\sqrt{ca}\le\dfrac{c+a}{2}\)
\(\Rightarrow\sqrt{a}+\sqrt{b}+\sqrt{c}+\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\le\dfrac{3\left(a+b+c\right)+3}{2}=\dfrac{3.3+3}{2}=6\)
Dấu "=" xảy ra khi \(a=b=c=1\)
Mà ta có: \(\sqrt{a}+\sqrt{b}+\sqrt{c}+\sqrt{ab}+\sqrt{bc}+\sqrt{ca}=6\)
\(\Rightarrow a=b=c=1\)
\(M=\dfrac{a^{30}+b^4+c^{1975}}{a^{30}+b^4+c^{2023}}=\dfrac{1^{30}+1^4+1^{1975}}{1^{30}+1^4+1^{2023}}=1\)
NV
Nguyễn Việt Lâm
Giáo viên
3 tháng 8 2021
\(\dfrac{\sqrt{ab+2c^2}}{\sqrt{1+ab-c^2}}=\dfrac{\sqrt{ab+2c^2}}{\sqrt{a^2+b^2+ab}}=\dfrac{ab+2c^2}{\sqrt{\left(a^2+b^2+ab\right)\left(ab+2c^2\right)}}\ge\dfrac{2\left(ab+2c^2\right)}{a^2+b^2+2ab+2c^2}\)
\(\ge\dfrac{2\left(ab+2c^2\right)}{a^2+b^2+a^2+b^2+2c^2}=\dfrac{ab+2c^2}{a^2+b^2+c^2}=ab+2c^2\)
Tương tự và cộng lại:
\(VT\ge ab+bc+ca+2\left(a^2+b^2+c^2\right)=2+ab+bc+ca\)
NV
Nguyễn Việt Lâm
Giáo viên
3 tháng 3 2021
\(N=\dfrac{\left(ab\right)^3+\left(bc\right)^3+\left(ca\right)^3}{\left(ab\right)\left(bc\right)\left(ca\right)}\)
Đặt \(\left(ab;bc;ca\right)=\left(x;y;z\right)\Rightarrow x+y+z=0\Rightarrow N=\dfrac{x^3+y^3+z^3}{xyz}\)
\(N=\dfrac{x^3+y^3+z^3-3xyz+3xyz}{xyz}=\dfrac{\dfrac{1}{2}\left(x+y+z\right)\left[\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\right]+3xyz}{xyz}=\dfrac{3xyz}{xyz}=3\)
30 tháng 12 2021
\(\dfrac{ab}{\sqrt{ab+2c}}=\dfrac{ab}{\sqrt{ab+\left(a+b+c\right)c}}=\dfrac{ab}{\sqrt{\left(a+c\right)\left(b+c\right)}}=ab\cdot\sqrt{\dfrac{1}{a+b}\cdot\dfrac{1}{b+c}}\le ab\cdot\dfrac{1}{2}\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}\right)=\dfrac{1}{2}\left(\dfrac{ab}{a+b}+\dfrac{ab}{b+c}\right)\)
CMTT: \(\dfrac{bc}{\sqrt{bc+2a}}\le\dfrac{1}{2}\left(\dfrac{bc}{a+b}+\dfrac{bc}{a+c}\right);\dfrac{ac}{\sqrt{ac+2b}}\le\dfrac{1}{2}\left(\dfrac{ac}{b+c}+\dfrac{ac}{b+a}\right)\)
\(\Leftrightarrow P\le\dfrac{1}{2}\left(\dfrac{ab}{c+a}+\dfrac{ab}{c+b}+\dfrac{bc}{b+a}+\dfrac{bc}{c+a}+\dfrac{ac}{b+c}+\dfrac{ac}{b+c}\right)\\ \Leftrightarrow P\le\dfrac{1}{2}\left[\dfrac{b\left(a+c\right)}{a+c}+\dfrac{a\left(b+c\right)}{b+c}+\dfrac{c\left(a+b\right)}{a+b}\right]=\dfrac{1}{2}\left(a+b+c\right)=1\)
Dấu \("="\Leftrightarrow a=b=c=\dfrac{2}{3}\)
Lời giải:
Từ \(ab+bc+ac=1\Rightarrow a^2+ab+bc+ac=a^2+1\)
\(\Leftrightarrow (a+b)(a+c)=a^2+1\)
Tương tự: \(\left\{\begin{matrix} b^2+1=(b+c)(b+a)\\ c^2+1=(c+a)(c+b)\end{matrix}\right.\)
Khi đó:
\(A=\frac{(b^2+bc)(c^2+ca)(a^2+ab)}{\sqrt{(a^4+a^2)(b^4+b^2)(c^4+c^2)}}\) \(=\frac{b(b+c)c(c+a)a(a+b)}{\sqrt{a^2b^2c^2(a^2+1)(b^2+1)(c^2+1)}}\)
\(=\frac{abc(a+b)(b+c)(c+a)}{abc\sqrt{(a+b)(a+c)(b+c)(b+a)(c+a)(c+b)}}\) \(=\frac{abc(a+b)(b+c)(c+a)}{abc(a+b)(b+c)(c+a)}=1\)
Vậy \(A=1\)