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Lời giải:
Áp dụng BĐT Cauchy-Schwarz:
\(\frac{1}{a+c}+\frac{1}{b+c}\geq \frac{4}{a+c+b+c}=\frac{4}{2c+a+b}\)
\(\Rightarrow \frac{ab}{a+c}+\frac{ab}{b+c}\ge \frac{4ab}{2c+a+b}\)
Tương tự: \(\frac{bc}{a+c}+\frac{bc}{a+b}\geq \frac{4bc}{2a+b+c}\); \(\frac{ca}{b+a}+\frac{ca}{b+c}\geq \frac{4ca}{2b+a+c}\)
Cộng 3 BĐT vừa thu được theo vế :
\(\Rightarrow \frac{ab}{a+c}+\frac{ab}{b+c}+\frac{bc}{a+c}+\frac{bc}{a+b}+\frac{ca}{b+a}+\frac{ca}{b+c}\geq 4P\)
\(\Leftrightarrow \frac{ab+bc}{a+c}+\frac{ab+ca}{b+c}+\frac{bc+ca}{a+b}\geq 4P\Leftrightarrow a+b+c\geq 4P\)
\(\Leftrightarrow 3\geq 4P\Leftrightarrow P\leq \frac{3}{4}\)
Vậy \(P_{\max}=\frac{3}{4}\Leftrightarrow a=b=c=1\)
\(\dfrac{ab}{\sqrt{ab+2c}}=\dfrac{ab}{\sqrt{ab+\left(a+b+c\right)c}}=\dfrac{ab}{\sqrt{\left(a+c\right)\left(b+c\right)}}=ab\cdot\sqrt{\dfrac{1}{a+b}\cdot\dfrac{1}{b+c}}\le ab\cdot\dfrac{1}{2}\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}\right)=\dfrac{1}{2}\left(\dfrac{ab}{a+b}+\dfrac{ab}{b+c}\right)\)
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 \("="\Leftrightarrow a=b=c=\dfrac{2}{3}\)
Lợi dụng Cauchy-Schwarz' inequality ta có:
\(\dfrac{ab}{\sqrt{ab+2c}}=\dfrac{ab}{\sqrt{ab+\left(a+b+c\right)c}}=\dfrac{ab}{\sqrt{ab+ac+bc+c^2}}\)
\(=\dfrac{ab}{\sqrt{\left(a+c\right)\left(b+c\right)}}\le\dfrac{1}{2}\left(\dfrac{ab}{a+c}+\dfrac{ab}{b+c}\right)\)
Tương tự ta cũng có:
\(\dfrac{bc}{\sqrt{bc+2a}}\le\dfrac{1}{2}\left(\dfrac{bc}{a+b}+\dfrac{bc}{a+c}\right);\dfrac{ca}{\sqrt{ca+2b}}\le\dfrac{1}{2}\left(\dfrac{ca}{a+b}+\dfrac{ca}{b+c}\right)\)
Cộng theo vế 3 BĐT trên ta có:
\(P\le\dfrac{1}{2}\left(\dfrac{ab+bc}{a+c}+\dfrac{bc+ca}{a+b}+\dfrac{ab+ca}{b+c}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{b\left(a+c\right)}{a+c}+\dfrac{c\left(a+b\right)}{a+b}+\dfrac{a\left(b+c\right)}{b+c}\right)\)
\(=\dfrac{1}{2}\left(a+b+c\right)=\dfrac{1}{2}\cdot2=1\)
Dấu "=" xảy ra khi \(a=b=c=\dfrac{2}{3}\)
Ta có P=\(\dfrac{ab}{\sqrt{ab+\left(a+b+c\right)c}}+\dfrac{bc}{\sqrt{bc+\left(a+b+c\right)a}}+\dfrac{ac}{\sqrt{ac+\left(a+b+c\right)b}}\)
=\(\dfrac{ab}{\sqrt{ab+ac+bc+c^2}}+\dfrac{bc}{\sqrt{bc+ac+ab+a^2}}+\dfrac{ac}{\sqrt{ac+ab+bc+b^2}}\)
=\(\dfrac{ab}{\sqrt{a\left(b+c\right)+c\left(b+c\right)}}+\dfrac{bc}{\sqrt{b\left(a+c\right)+a\left(a+c\right)}}+\dfrac{ac}{\sqrt{c\left(a+b\right)+b\left(a+b\right)}}\)
=\(\dfrac{ab}{\sqrt{\left(a+c\right)\left(b+c\right)}}+\dfrac{bc}{\sqrt{\left(b+a\right)\left(c+a\right)}}+\dfrac{ac}{\sqrt{\left(a+b\right)\left(c+b\right)}}\)
áp dụng bđt Cói ta có:
\(\sqrt{\left(a+c\right)\left(b+c\right)}\)\(\le\)\(\dfrac{2+c}{2}=1+\dfrac{c}{2}\)
\(\sqrt{\left(b+á\right)\left(c+a\right)}\)
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Câu hỏi của Tuyển Trần Thị - Toán lớp 9 | Học trực tuyến
Áp dụng bất đẳng thức Cauchy-Schwarz:
\(\dfrac{1}{a}+\dfrac{2}{b}+\dfrac{3}{c}=\dfrac{1}{a}+\dfrac{4}{2b}+\dfrac{9}{3c}\ge\dfrac{\left(1+2+3\right)^2}{a+2b+3c}=\dfrac{36}{a+2b+3c}\)
\(\dfrac{2}{a}+\dfrac{3}{b}+\dfrac{1}{c}=\dfrac{4}{2a}+\dfrac{9}{3b}+\dfrac{1}{c}\ge\dfrac{\left(2+3+1\right)^2}{2a+3b+c}=\dfrac{36}{2a+3b+c}\)
\(\dfrac{3}{a}+\dfrac{1}{b}+\dfrac{2}{c}=\dfrac{9}{3a}+\dfrac{1}{b}+\dfrac{4}{2c}\ge\dfrac{\left(3+1+2\right)^2}{3a+b+2c}=\dfrac{36}{3a+2b+c}\)
Cộng theo vế: \(6\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge36F\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge6F\)
Mặt khác: \(ab+bc+ac=3abc\Leftrightarrow\dfrac{ab+bc+ac}{abc}=3\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=3\)
\(\Rightarrow18\ge36F\Leftrightarrow F\le\dfrac{1}{2}\)
Dấu "=" xảy ra khi: \(a=b=c=1\)
b, \(\frac{a^3}{b+2c}+\frac{b^3}{c+2a}+\frac{c^3}{a+2b}\ge1\)
\(\frac{a^4}{ab+2ac}+\frac{b^4}{bc+2ab}+\frac{c^4}{ac+2bc}\ge\frac{\left(a^2+b^2+c^2\right)^2}{ab+bc+ac+2ac+2ab+2bc}\)( Bunhia dạng phân thức )
mà \(a^2+b^2+c^2\ge ab+bc+ac\)
\(=\frac{\left(ab+bc+ac\right)^2}{3+2\left(ab+ac+bc\right)}=\frac{9}{3+6}=1\)( đpcm )
1.
Điều kiện x \ge \dfrac14x≥41.
Phương trình tương đương với \left(\sqrt2.\sqrt{2x^2+x+1}-2\right)-\left(\sqrt{4x-1}-1\right)+2x^2+3x-2 = 0(2.2x2+x+1−2)−(4x−1−1)+2x2+3x−2=0 \Leftrightarrow \dfrac{4x^2+2x-2}{\sqrt2.\sqrt{2x^2+x+1}+2} - \dfrac{4x-2}{\sqrt{4x-1}+1} + (x+2)(2x-1) = 0⇔2.2x2+x+1+24x2+2x−2−4x−1+14x−2+(x+2)(2x−1)=0\\ \Leftrightarrow (2x-1)\left(\dfrac{2(x+1)}{\sqrt2 \sqrt{2x^2+x+1}+2} - \dfrac2{\sqrt{4x-1}+1} + x + 2\right) = 0⇔(2x−1)(22x2+x+1+22(x+1)−4x−1+12+x+2)=0
\Leftrightarrow \left[\begin{aligned} & x =\dfrac12\\ & \dfrac{2(x+1)}{\sqrt2 \sqrt{2x^2+x+1}+2} - \dfrac2{\sqrt{4x-1}+1} + x + 2 = 0\\ \end{aligned}\right.⇔⎣⎢⎢⎢⎡x=2122x2+x+1+22(x+1)−4x−1+12+x+2=0
Với x \ge \dfrac14x≥41 ta có:
\dfrac{2(x+1)}{\sqrt2 \sqrt{2x^2+x+1}+2} > 022x2+x+1+22(x+1)>0
- \dfrac2{\sqrt{4x-1}+1} \ge -2−4x−1+12≥−2
x + 2 > 2x+2>2.
Suy ra \dfrac{2(x+1)}{\sqrt2 \sqrt{2x^2+x+1}+2} - \dfrac2{\sqrt{4x-1}+1} + x + 2 > 022x2+x+1+22(x+1)−4x−1+12+x+2>0.
Vậy phương trình có nghiệm duy nhất x = \dfrac12.x=21.
2.
Đặt P = \dfrac{a^3}{b+2c} + \dfrac{b^3}{c+2a} + \dfrac{c^3}{a+2b}P=b+2ca3+c+2ab3+a+2bc3
Áp dụng bất đẳng thức Cauchy cho hai số dương \dfrac{9a^3}{b + 2c}b+2c9a3 và (b+2c)a(b+2c)a ta có
\dfrac{9a^3}{b+2c} + (b+2c)a \ge 6a^2b+2c9a3+(b+2c)a≥6a2.
Tương tự \dfrac{9b^3}{c+2a} + (c+2a)b \ge 6b^2c+2a9b3+(c+2a)b≥6b2, \dfrac{9c^3}{a+2b} + (a+2b)c \ge 6c^2a+2b9c3+(a+2b)c≥6c2.
Cộng các vế ta có 9P + 3(ab+bc+ca) \ge 6(a^2+b^2+c^2)9P+3(ab+bc+ca)≥6(a2+b2+c2).
Mà a^2+b^2+c^2 \ge ab+bc+ca = 4a2+b2+c2≥ab+bc+ca=4 nên P \ge 1P≥1 (ta có đpcm).
https://hoc24.vn/hoi-dap/question/562943.html
Em xem ở đây nhé.
\(M=\dfrac{1}{\dfrac{c}{a}+\dfrac{2a}{b}+3}+\dfrac{1}{\dfrac{a}{b}+\dfrac{2b}{c}+3}+\dfrac{1}{\dfrac{b}{c}+\dfrac{2c}{a}+3}\)
\(đặt\left(\dfrac{a}{b};\dfrac{b}{c};\dfrac{c}{a}\right)=\left(x;y;z\right)\Rightarrow xyz=1\left(x;y;z>0\right)\)
\(M=\dfrac{1}{z+2x+3}+\dfrac{1}{x+2y+3}+\dfrac{1}{y+2z+3}\)
\(ta\) \(đi\) \(cminh:A\le\dfrac{1}{2}\)
có:
\(\dfrac{1}{z+2x+3}\le\dfrac{1}{6}\Leftrightarrow z+2x+3\ge6\Leftrightarrow2x+z\ge3\)
\(\dfrac{1}{x+2y+3}\le\dfrac{1}{6}\Leftrightarrow x+2y\ge3\)
\(\dfrac{1}{y+2z+3}\le\dfrac{1}{6}\Rightarrow y+2z\ge3\)
\(cộng\) \(vế\Rightarrow2x+z+2y+x+2z+y\ge9\Leftrightarrow x+y+z\ge3\left(đúng\right)\)
\(do:x+y+z\ge3\sqrt[3]{xyz}=3\)
\(\Rightarrow A\le\dfrac{1}{2}dấu"="\Leftrightarrow x=y=z=1\Rightarrow a=b=c\)
giúp bài nghiệm nguyên lun đk ạ