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Em nghĩ đề là a chứ không phải 2a ;v
\(P=\dfrac{a}{\sqrt{1+a^2}}+\dfrac{b}{\sqrt{1+b^2}}+\dfrac{c}{\sqrt{1+c^2}}\\ =\dfrac{a}{\sqrt{ab+bc+ac+a^2}}+\dfrac{b}{\sqrt{ab+bc+ac+b^2}}+\dfrac{c}{\sqrt{ab+bc+ac+c^2}}\\ =\dfrac{a}{\sqrt{\left(a+b\right)\left(a+c\right)}}+\dfrac{b}{\sqrt{\left(a+b\right)\left(b+c\right)}}+\dfrac{c}{\sqrt{\left(a+c\right)\left(b+c\right)}}\\ \le\left(\dfrac{a}{2\left(a+b\right)}+\dfrac{a}{2\left(a+c\right)}\right)+\left(\dfrac{b}{2\left(a+b\right)}+\dfrac{b}{2\left(b+c\right)}\right)+\left(\dfrac{c}{2\left(a+c\right)}+\dfrac{c}{2\left(b+c\right)}\right)\)
\(=\dfrac{2\left(a+b+c\right)}{8\left(a+b+c\right)}=\dfrac{1}{4}\)
Áp dụng bđt : \(\dfrac{1}{xy}\le\dfrac{\dfrac{1}{x^2}+\dfrac{1}{y^2}}{2}\)
Dấu "=" xảy ra khi a=b=c=1/căn 3
Dự đoán điểm rơi b=c=ka. Ta có:
\(P=\dfrac{2a}{\sqrt{\left(a+b\right)\left(a+c\right)}}+\dfrac{b}{\sqrt{\left(b+c\right)\left(b+a\right)}}+\dfrac{c}{\sqrt{\left(c+a\right)\left(c+b\right)}}\)
Áp dụng BĐT AM-GM: \(\dfrac{2a}{\sqrt{\left(a+b\right)\left(a+c\right)}}\le\dfrac{a}{a+b}+\dfrac{a}{a+c}\)
\(\dfrac{b}{\sqrt{\left(b+c\right)\left(b+a\right)}}=\dfrac{b.\sqrt{\dfrac{2k}{k+1}}}{\sqrt{\left(b+c\right).\dfrac{2k\left(a+b\right)}{k+1}}}\le\dfrac{b}{2}\sqrt{\dfrac{2k}{k+1}}.\left(\dfrac{1}{b+c}+\dfrac{\left(k+1\right)}{2k\left(a+b\right)}\right)\)
\(\dfrac{c}{\sqrt{\left(c+a\right)\left(c+b\right)}}\le\dfrac{c}{2}.\sqrt{\dfrac{2k}{k+1}}\left(\dfrac{1}{b+c}+\dfrac{k+1}{2k\left(a+c\right)}\right)\)
\(\Rightarrow VT\le\dfrac{a}{a+b}+\sqrt{\dfrac{k+1}{8k}}.\dfrac{b}{a+b}+\dfrac{a}{a+c}+\sqrt{\dfrac{k+1}{8k}}.\dfrac{c}{a+c}+\sqrt{\dfrac{k}{2k+2}}\)
Tìm k sao cho \(\sqrt{\dfrac{k+1}{8k}}=1\Rightarrow k=\dfrac{1}{7}\)
Do đó trình bày lại bài toán ngắn gọn như sau:
Áp dụng BĐT AM-GM:
\(VT=\dfrac{2a}{\sqrt{\left(a+b\right)\left(a+c\right)}}+\dfrac{2b}{\sqrt{4\left(b+c\right).\left(b+a\right)}}+\dfrac{2c}{\sqrt{4\left(b+c\right).\left(a+b\right)}}\)
\(\le\dfrac{a}{a+b}+\dfrac{a}{a+c}+\dfrac{b}{4\left(b+c\right)}+\dfrac{b}{a+b}+\dfrac{c}{4\left(b+c\right)}+\dfrac{c}{a+c}\)
\(=1+1+\dfrac{1}{4}=\dfrac{9}{4}\)
Dấu = xảy ra khi \(a=7b=7c=\dfrac{7}{\sqrt{15}}\)
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Câu hỏi của Tuyển Trần Thị - Toán lớp 9 | Học trực tuyến
\(a^2-ab+b^2=\dfrac{1}{4}\left(a+b\right)^2+\dfrac{3}{4}\left(a-b\right)^2\ge\dfrac{1}{4}\left(a+b\right)^2\)
\(\Rightarrow P\le\dfrac{2}{a+b}+\dfrac{2}{b+c}+\dfrac{2}{c+a}\le\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=3\)
Dấu "=" xảy ra khi \(a=b=c=1\)
\(Q=\dfrac{2a}{\sqrt{a^2+ab+bc+ca}}+\dfrac{b}{\sqrt{b^2+ab+bc+ca}}+\dfrac{c}{\sqrt{c^2+ab+bc+ca}}\)
\(=\dfrac{2a}{\sqrt{\left(a+b\right)\left(a+c\right)}}+\dfrac{b}{\sqrt{\left(a+b\right)\left(b+c\right)}}+\dfrac{c}{\sqrt{\left(a+c\right)\left(b+c\right)}}\)
\(=\sqrt{\dfrac{2a}{a+b}.\dfrac{2a}{a+c}}+\sqrt{\dfrac{2b}{a+b}.\dfrac{b}{2\left(b+c\right)}}+\sqrt{\dfrac{2c}{a+c}.\dfrac{c}{2\left(b+c\right)}}\)
\(\le\dfrac{1}{2}\left(\dfrac{2a}{a+b}+\dfrac{2a}{a+c}+\dfrac{2b}{a+b}+\dfrac{b}{2\left(b+c\right)}+\dfrac{2c}{a+c}+\dfrac{c}{2\left(b+c\right)}\right)\)
\(=\dfrac{9}{4}\)
Dấu "=" xảy ra khi \(\left(a;b;c\right)=\left(\dfrac{7}{\sqrt{15}};\dfrac{1}{\sqrt{15}};\dfrac{1}{\sqrt{15}}\right)\)
\(P=\sqrt{\dfrac{ab}{c+ab}}+\sqrt{\dfrac{bc}{a+bc}}+\sqrt{\dfrac{ca}{b+ca}}\)
\(P=\sqrt{\dfrac{ab}{c\left(a+b+c\right)+ab}}+\sqrt{\dfrac{bc}{a\left(a+b+c\right)+bc}}+\sqrt{\dfrac{ca}{b\left(a+b+c\right)+ca}}\)
\(P=\sqrt{\dfrac{ab}{ac+bc+c^2+ab}}+\sqrt{\dfrac{bc}{a^2+ab+ac+bc}}+\sqrt{\dfrac{ca}{ab+b^2+bc+ca}}\)
\(P=\sqrt{\dfrac{ab}{\left(a+c\right)\left(b+c\right)}}+\sqrt{\dfrac{bc}{\left(a+b\right)\left(a+c\right)}}+\sqrt{\dfrac{ca}{\left(a+b\right)\left(b+c\right)}}\)
Áp dụng bất đẳng thức Cauchy cho 2 bộ số thực không âm
\(\Rightarrow\left\{{}\begin{matrix}\sqrt{\dfrac{ab}{\left(a+c\right)\left(b+c\right)}}\le\dfrac{\dfrac{a}{a+c}+\dfrac{b}{b+c}}{2}\\\sqrt{\dfrac{bc}{\left(a+b\right)\left(a+c\right)}}\le\dfrac{\dfrac{b}{a+b}+\dfrac{c}{a+c}}{2}\\\sqrt{\dfrac{ca}{\left(a+b\right)\left(b+c\right)}}\le\dfrac{\dfrac{a}{a+b}+\dfrac{c}{b+c}}{2}\end{matrix}\right.\)
\(\Rightarrow VT\le\dfrac{\left(\dfrac{a}{a+c}+\dfrac{c}{a+c}\right)+\left(\dfrac{b}{b+c}+\dfrac{c}{b+c}\right)+\left(\dfrac{b}{a+b}+\dfrac{a}{a+b}\right)}{2}\)
\(\Rightarrow VT\le\dfrac{\dfrac{a+c}{a+c}+\dfrac{b+c}{b+c}+\dfrac{a+b}{a+b}}{2}=\dfrac{3}{2}\)
\(\Rightarrow P\le\dfrac{3}{2}\)
Vậy \(P_{max}=\dfrac{3}{2}\)
Dấu " = " xảy ra khi \(a=b=c=\dfrac{1}{3}\)
Đặt \(\left(\dfrac{1}{a};\dfrac{1}{b};\dfrac{1}{c}\right)\rightarrow\left(x;y;z\right)\)\(\Rightarrow\left\{{}\begin{matrix}x,y,z>0\\\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{xz}=1\end{matrix}\right.\)\(\Rightarrow x+y+z=xyz\)
\(\Rightarrow P=xy+yz+xz-\sqrt{x^2+1}-\sqrt{y^2+1}-\sqrt{z^2+1}\)
Khi \(a=b=c=\frac{1}{\sqrt{3}}\Rightarrow x=y=z=\sqrt{3}\Rightarrow P=3\)
Ta sẽ chứng minh \(P=3\) là giá tri nhỏ nhất của \(P\)
\(\Rightarrow BDT\Leftrightarrow xy+yz+xz-3\ge\sqrt{x^2+1}+\sqrt{y^2+1}+\sqrt{z^2+1}\)
Ta có BĐT \(\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}\ge\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{xz}=1\)
\(\Leftrightarrow x^2y^2+y^2z^2+z^2x^2\ge x^2y^2z^2\)
\(\Leftrightarrow\left(xy+yz+xz\right)^2\ge x^2y^2z^2+2xyz\left(x+y+z\right)\)\(=3\left(x+y+z\right)^2\)
Xét \(VT^2=\left(xy+yz+xz-3\right)^2=\left(xy+yz+xz\right)^2-6\left(xy+yz+xz\right)+9\)
\(\ge3\left(x+y+z\right)^2-6\left(xy+yz+xz\right)+9\)\(=3\left(x^2+y^2+z^2\right)+9\left(1\right)\)
Và \(VP^2\le\left(1+1+1\right)\left(x^2+y^2+z^2+3\right)=3\left(x^2+y^2+z^2\right)+9\left(2\right)\)
Từ \(\left(1\right);\left(2\right)\) ta có ĐPCM. Vậy \(P_{min}=3\Rightarrow a=b=c=\frac{1}{\sqrt{3}}\)
\(\dfrac{ab}{\sqrt{ab+2c}}=\dfrac{ab}{\sqrt{ab+\left(a+b+c\right)c}}=\dfrac{ab}{\sqrt{\left(a+c\right)\left(b+c\right)}}=ab\cdot\sqrt{\dfrac{1}{a+b}\cdot\dfrac{1}{b+c}}\le ab\cdot\dfrac{1}{2}\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}\right)=\dfrac{1}{2}\left(\dfrac{ab}{a+b}+\dfrac{ab}{b+c}\right)\)
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 \("="\Leftrightarrow a=b=c=\dfrac{2}{3}\)
Bài toán quy về 2 bài toán nhỏ hơn!
Cho các số dương ab + bc +ca = 1.
a) Tìm Max: \(M=\frac{2a}{\sqrt{1+a^2}}+\frac{b}{\sqrt{1+b^2}}+\frac{c}{\sqrt{1+c^2}}\)
(Lời giải tại: Câu hỏi của Nguyễn Linh Chi. Bài làm của anh Thắng, trong lời giải có phần giống với đề bên trên.)
b) Tìm Min: \(N=a^2+28b^2+28c^2\)
Có: \(N=\frac{1}{4}\left(2a-7b-7c\right)^2+\frac{63}{4}\left(b-c\right)^2+7\left(ab+bc+ca\right)\ge7\left(ab+bc+ca\right)=7\)
Từ đó tìm được \(P\le\frac{9}{4}-7=-\frac{19}{4}\)
Đẳng thức xảy ra khi \(a=\frac{7}{\sqrt{15}};b=c=\frac{1}{\sqrt{15}}\)
Với ab + bc + ca = 1 và áp dụng BĐT AM - GM, ta được:
\(\frac{2a}{\sqrt{1+a^2}}+\frac{b}{\sqrt{1+b^2}}+\frac{c}{\sqrt{1+c^2}}\)\(\frac{2a}{\sqrt{a^2+ab+bc+ca}}+\frac{b}{\sqrt{b^2+ab+bc+ca}}+\frac{c}{\sqrt{c^2+ab+bc+ca}}\)
\(=\frac{2a}{\sqrt{\left(a+b\right)\left(a+c\right)}}+\frac{b}{\sqrt{\left(b+a\right)\left(b+c\right)}}+\frac{c}{\sqrt{\left(c+a\right)\left(c+b\right)}}\)
\(=\sqrt{\frac{2a}{a+b}.\frac{2a}{a+c}}+\sqrt{\frac{2b}{a+b}.\frac{b}{2\left(b+c\right)}}+\sqrt{\frac{2c}{a+c}.\frac{c}{2\left(b+c\right)}}\)
\(\le\frac{\frac{2a}{a+b}+\frac{2a}{a+c}}{2}+\frac{\frac{2b}{a+b}+\frac{b}{2\left(b+c\right)}}{2}+\frac{\frac{2c}{a+c}+\frac{c}{2\left(b+c\right)}}{2}\)
\(=\frac{\frac{2\left(a+b\right)}{a+b}+\frac{2\left(a+c\right)}{a+c}+\frac{b+c}{2\left(b+c\right)}}{2}=\frac{2+2+\frac{1}{2}}{2}=\frac{9}{4}\)(*)
Mặt khác, cũng theo AM - GM, ta có:
\(\frac{a^2}{2}+\frac{49b^2}{2}\ge2\sqrt{\frac{a^2}{2}.\frac{49b^2}{2}}=7ab\)(1)
\(\frac{a^2}{2}+\frac{49c^2}{2}\ge2\sqrt{\frac{a^2}{2}.\frac{49c^2}{2}}=7ac\)(2)
\(\frac{7}{2}\left(b^2+c^2\right)\ge\frac{7}{2}.2\sqrt{b^2c^2}=7bc\)(3)
Cộng theo từng vế của 3 BĐT (1), (2), (3), ta được:
\(\frac{2a^2+56b^2+56c^2}{2}\ge7\left(ab+bc+ca\right)=7\)
hay \(a^2+28b^2+28c^2\ge7\)(**)
Từ (*) và (**) suy ra \(P=\frac{2a}{\sqrt{1+a^2}}+\frac{b}{\sqrt{1+b^2}}+\frac{c}{\sqrt{1+c^2}}-a^2-28b^2-28c^2\)
\(\le\frac{9}{4}-7=\frac{-19}{4}\)
Đẳng thức xảy ra khi \(a=\frac{7}{\sqrt{15}};b=c=\frac{1}{\sqrt{15}}\)
Lời giải:
Áp dụng BĐT AM-GM:
\(P=\frac{2a}{\sqrt{a^2+ab+bc+ac}}+\frac{b}{\sqrt{b^2+ab+bc+ac}}+\frac{c}{\sqrt{c^2+ab+bc+ac}}\\ =\frac{2a}{\sqrt{(a+b)(a+c)}}+\frac{b}{\sqrt{(b+c)(b+a)}}+\frac{c}{\sqrt{(c+a)(c+b)}}\)
\(\leq \frac{a}{a+b}+\frac{a}{a+c}+\frac{b}{4(b+c)}+\frac{b}{b+a}+\frac{c}{4(c+b)}+\frac{c}{c+a}\)
\(=(\frac{a}{a+b}+\frac{b}{b+a})+(\frac{a}{a+c}+\frac{c}{a+c})+\frac{1}{4}(\frac{b}{b+c}+\frac{c}{b+c})=1+1+\frac{1}{4}=\frac{9}{4}\)
Vậy $P_{\max}=\frac{9}{4}$