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Áp dụng bđt Schwarz ta có:
\(P=\dfrac{a^4}{2ab+3ac}+\dfrac{b^4}{2cb+3ab}+\dfrac{c^4}{2ac+3bc}\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{5\left(ab+bc+ca\right)}\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{5\left(a^2+b^2+c^2\right)}=\dfrac{1}{5}\).
Đẳng thức xảy ra khi và chỉ khi \(a=b=c=\dfrac{\sqrt{3}}{3}\).
Áp dụng BĐT Holder:
\(\left(\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}\right)^2\left[a^2\left(b+c\right)^2+b^2\left(c+a\right)^2+c^2\left(a+b\right)^2\right]\ge\left(a^2+b^2+c^2\right)^3\)
Mặt khác:
\(\left(a^2+b^2+c^2\right)^2\ge3\left(a^2b^2+b^2c^2+c^2a^2\right)\ge\dfrac{3}{2}\left(a^2b^2+b^2c^2+c^2a^2+abc\left(a+b+c\right)\right)\)
\(\Rightarrow\left(a^2+b^2+c^2\right)^2\ge\dfrac{3}{4}\left[a^2\left(b+c\right)^2+b^2\left(c+a\right)^2+c^2\left(a+b\right)^2\right]\)
\(\Rightarrow\left(\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}\right)^2\ge\dfrac{3}{4}\left(a^2+b^2+c^2\right)\)
\(\Rightarrow\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}\ge\dfrac{\sqrt{3}}{2}\sqrt{a^2+b^2+c^2}\)
\(\Rightarrow P\ge\dfrac{\sqrt{3}}{2}\sqrt{a^2+b^2+c^2}+\dfrac{4}{\sqrt{a^2+b^2+c^2+1}}\)
Đặt \(\sqrt{\dfrac{a^2+b^2+c^2}{3}}=x>0\)
\(\Rightarrow P\ge\dfrac{3x}{2}+\dfrac{4}{\sqrt{3x^2+1}}\)
Ta sẽ chứng minh \(P\ge\dfrac{7}{2}\)
Thật vậy, với \(x\ge\dfrac{7}{3}\Rightarrow P>\dfrac{3x}{2}\ge\dfrac{7}{2}\) (đúng)
Với \(0< x\le\dfrac{7}{3}\) ta cần chứng minh:
\(\dfrac{3x}{2}+\dfrac{4}{\sqrt{3x^2+1}}\ge\dfrac{7}{2}\Leftrightarrow\dfrac{4}{\sqrt{3x^2+1}}\ge\dfrac{7-3x}{2}\)
\(\Leftrightarrow64\ge\left(7-3x\right)^2\left(3x^2+1\right)\)
\(\Leftrightarrow3\left(x-1\right)^2\left(-9x^2+24x+5\right)\ge0\)
\(\Leftrightarrow\left(x-1\right)^2\left[3x\left(7-3x\right)+3x+5\right]\ge0\) (đúng)
Vậy \(P_{min}=\dfrac{7}{2}\) khi \(x=1\) hay \(a=b=c=1\)
a. cos2x = 1-sin2x
b. cos2x = 2cos2x - 1
c. 2cosx.cos2x = 1 + cos2x + cos3x
=> 2cosx.cos2x = 2cos2x + 4cos3x - 3cosx
=> cosx(2.(2cos2x - 1) - 2cosx - 4cos2x +3) = 0
=> cosx( -2cosx + 1) = 0
=> cosx=0 hoặc cosx = -1/2
Bạn tham khảo:
Câu hỏi của Thảo Nguyễn Phương - Toán lớp 11 | Học trực tuyến
16.
\(y'=\frac{\left(cos2x\right)'}{2\sqrt{cos2x}}=\frac{-2sin2x}{2\sqrt{cos2x}}=-\frac{sin2x}{\sqrt{cos2x}}\)
17.
\(y'=4x^3-\frac{1}{x^2}-\frac{1}{2\sqrt{x}}\)
18.
\(y'=3x^2-2x\)
\(y'\left(-2\right)=16;y\left(-2\right)=-12\)
Pttt: \(y=16\left(x+2\right)-12\Leftrightarrow y=16x+20\)
19.
\(y'=-\frac{1}{x^2}=-x^{-2}\)
\(y''=2x^{-3}=\frac{2}{x^3}\)
20.
\(\left(cotx\right)'=-\frac{1}{sin^2x}\)
21.
\(y'=1+\frac{4}{x^2}=\frac{x^2+4}{x^2}\)
22.
\(lim\left(3^n\right)=+\infty\)
11.
\(\lim\limits_{x\rightarrow1^+}\frac{-2x+1}{x-1}=\frac{-1}{0}=-\infty\)
12.
\(y=cotx\Rightarrow y'=-\frac{1}{sin^2x}\)
13.
\(y'=2020\left(x^3-2x^2\right)^{2019}.\left(x^3-2x^2\right)'=2020\left(x^3-2x^2\right)^{2019}\left(3x^2-4x\right)\)
14.
\(y'=\frac{\left(4x^2+3x+1\right)'}{2\sqrt{4x^2+3x+1}}=\frac{8x+3}{2\sqrt{4x^2+3x+1}}\)
15.
\(y'=4\left(x-5\right)^3\)
Đặt \(\left(a-b;b-c;c-a\right)=\left(x;y;z\right)\Rightarrow\left\{{}\begin{matrix}x+y+z=0\\x^2+y^2+z^2=12\end{matrix}\right.\)
\(\Rightarrow xy+yz+zx=-6\Rightarrow yz=-6-x\left(y+z\right)=x^2-6\)
Ta cũng có: \(12=x^2+y^2+z^2\ge x^2+\frac{1}{2}\left(y+z\right)^2=\frac{3}{2}x^2\Rightarrow-2\sqrt{2}\le x\le2\sqrt{2}\)
\(P=xyz=x\left(x^2-6\right)=x^3-6x\)
Xét hàm \(f\left(x\right)=x^3-6x\) trên \(\left[-2\sqrt{2};2\sqrt{2}\right]\Rightarrow f'\left(x\right)=3x^2-6=0\Rightarrow x=\pm\sqrt{2}\)
\(f\left(-2\sqrt{2}\right)=-4\sqrt{2}\) ; \(f\left(-\sqrt{2}\right)=4\sqrt{2}\) ; \(f\left(\sqrt{2}\right)=-4\sqrt{2}\) ; \(f\left(2\sqrt{2}\right)=4\sqrt{2}\)
\(\Rightarrow P_{max}=4\sqrt{2}\) ; \(P_{min}=-4\sqrt{2}\)
\(\dfrac{a^5}{b^3+c^2}+\dfrac{b^3+c^2}{4}+\dfrac{a^4}{2}\ge3\sqrt[3]{\dfrac{a^9.\left(b^3+c^2\right)}{8\left(b^3+c^2\right)}}=\dfrac{3a^3}{2}\)
Tương tự và cộng lại:
\(\Rightarrow M-\dfrac{a^4+b^4+c^4}{2}+\dfrac{a^3+b^3+c^3}{4}+\dfrac{a^2+b^2+c^2}{4}\ge\dfrac{3}{2}\left(a^3+b^3+c^3\right)\)
\(\Rightarrow M\ge\dfrac{a^4+b^4+c^4}{2}+\dfrac{5}{4}\left(a^3+b^3+c^3\right)-\dfrac{3}{4}\)
Mặt khác ta có:
\(\dfrac{1}{2}\left(a^4+b^4+c^4\right)\ge\dfrac{1}{6}\left(a^2+b^2+c^2\right)^2=\dfrac{3}{2}\)
\(\left(a^3+a^3+1\right)+\left(b^3+b^3+1\right)+\left(c^3+c^3+1\right)\ge3\left(a^2+b^2+c^2\right)=9\)
\(\Rightarrow2\left(a^3+b^3+c^3\right)+3\ge9\Rightarrow a^3+b^3+c^3\ge3\)
\(\Rightarrow M\ge\dfrac{3}{2}+\dfrac{15}{4}-\dfrac{3}{4}=...\)