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1.
\(6=\frac{\sqrt{2}^2}{x}+\frac{\sqrt{3}^2}{y}\ge\frac{\left(\sqrt{2}+\sqrt{3}\right)^2}{x+y}=\frac{5+2\sqrt{6}}{x+y}\)
\(\Rightarrow x+y\ge\frac{5+2\sqrt{6}}{6}\)
Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}\frac{x}{\sqrt{2}}=\frac{y}{\sqrt{3}}\\x+y=\frac{5+2\sqrt{6}}{6}\end{matrix}\right.\)
Bạn tự giải hệ tìm điểm rơi nếu thích, số xấu quá
2.
\(VT\ge\sqrt{\left(x+y+z\right)^2+\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2}\ge\sqrt{\left(x+y+z\right)^2+\frac{81}{\left(x+y+z\right)^2}}\)
Đặt \(x+y+z=t\Rightarrow0< t\le1\)
\(VT\ge\sqrt{t^2+\frac{81}{t^2}}=\sqrt{t^2+\frac{1}{t^2}+\frac{80}{t^2}}\ge\sqrt{2\sqrt{\frac{t^2}{t^2}}+\frac{80}{1^2}}=\sqrt{82}\)
Dấu "=" xảy ra khi \(x=y=z=\frac{1}{3}\)
3.
\(\frac{a^2}{b^5}+\frac{a^2}{b^5}+\frac{a^2}{b^5}+\frac{1}{a^3}+\frac{1}{a^3}\ge5\sqrt[5]{\frac{a^6}{b^{15}.a^6}}=\frac{5}{b^3}\)
Tương tự: \(\frac{3b^2}{c^5}+\frac{2}{b^3}\ge\frac{5}{a^3}\) ; \(\frac{3c^2}{d^5}+\frac{2}{c^3}\ge\frac{5}{d^3}\) ; \(\frac{3d^2}{a^5}+\frac{2}{d^2}\ge\frac{5}{a^3}\)
Cộng vế với vế và rút gọn ta được: \(3VT\ge3VP\)
Dấu "=" xảy ra khi và chỉ khi \(a=b=c=d=1\)
4.
ĐKXĐ: \(-2\le x\le2\)
\(y^2=\left(x+\sqrt{4-x^2}\right)^2\le2\left(x^2+4-x^2\right)=8\)
\(\Rightarrow y\le2\sqrt{2}\Rightarrow y_{max}=2\sqrt{2}\) khi \(x=\sqrt{2}\)
Mặt khác do \(\left\{{}\begin{matrix}x\ge-2\\\sqrt{4-x^2}\ge0\end{matrix}\right.\) \(\Rightarrow x+\sqrt{4-x^2}\ge-2\)
\(y_{min}=-2\) khi \(x=-2\)
1)
\(2a+\frac{4}{a}+\frac{16}{a+2}=\left(a+\frac{4}{a}\right)+\left[\left(a+2\right)+\frac{16}{a+2}\right]-2\ge4+8-2=10\)
Dấu "=" xảy ra khi a=2
2)
\(\hept{\begin{cases}\sqrt{a\left(1-4a\right)}=\frac{1}{2}\sqrt{4a\left(1-4a\right)}\le\frac{1}{2}\cdot\frac{4a+1-4a}{2}=\frac{1}{4}\\\sqrt{b\left(1-4b\right)}=\frac{1}{2}\sqrt{4\left(1-4a\right)}\le\frac{1}{2}\cdot\frac{4b+1-4b}{2}=\frac{1}{4}\\\sqrt{c\left(1-4c\right)}=\frac{1}{2}\sqrt{4c\left(1-4c\right)}\le\frac{1}{2}\cdot\frac{4c+1-4c}{2}=\frac{1}{4}\end{cases}}\)
\(\Rightarrow\sqrt{a\left(1-4a\right)}+\sqrt{b\left(1-4b\right)}+\sqrt{c\left(1-4c\right)}\le\frac{3}{4}\)
Dấu "=" xảy ra khi \(a=b=c=\frac{1}{8}\)
\(a^3+b^3\ge ab\left(a+b\right)\)
\(\Leftrightarrow\left(a+b\right)\left(a^2-ab+b^2\right)-ab\left(a+b\right)\ge0\)
\(\Leftrightarrow\left(a+b\right)\left(a-b\right)^2\ge0\) ( đúng )
Dấu "=" \(\Leftrightarrow a=b\)
a) Áp dụng BĐT trên ta có:
\(\Sigma\left(\frac{1}{a^3+b^3+abc}\right)\le\Sigma\left(\frac{1}{ab\left(a+b\right)+abc}\right)=\Sigma\left[\frac{1}{ab}\cdot\left(\frac{1}{a+b+c}\right)\right]=\frac{1}{a+b+c}\cdot\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)=\frac{a+b+c}{\left(a+b+c\right)\cdot abc}=\frac{1}{abc}\)
Dấu "=" khi \(a=b=c\)
b) \(\Sigma\left(\frac{1}{a^3+b^3+1}\right)\le\Sigma\left(\frac{1}{ab\left(a+b\right)+abc}\right)=\Sigma\left[\frac{1}{ab}\cdot\left(\frac{1}{a+b+c}\right)\right]=\frac{1}{abc}=1\)
Dấu "=" khi \(a=b=c=1\)
c) \(\Sigma\left(\frac{1}{a+b+1}\right)\le\Sigma\left(\frac{1}{\sqrt[3]{ab}\left(\sqrt[3]{a}+\sqrt[3]{b}\right)+\sqrt[3]{abc}}\right)=\Sigma\left[\frac{1}{\sqrt[3]{ab}\left(\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}\right)}\right]\)
\(=\frac{1}{\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}}\cdot\left(\frac{1}{\sqrt[3]{ab}}+\frac{1}{\sqrt[3]{bc}}+\frac{1}{\sqrt[3]{ca}}\right)=\frac{\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}}{\left(\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}\right)\cdot\sqrt[3]{abc}}=\frac{1}{\sqrt[3]{abc}}=1\)
Dấu "=" khi \(a=b=c=1\)
1) ĐK: \(\frac{x+1}{x}>0\Leftrightarrow\left[\begin{array}{nghiempt}x>0\\x< -1\end{array}\right.\)
Đặt \(t=\sqrt{\frac{x+1}{x}}\left(t>0\right)\) , bất pt đã cho trở thành:
\(\frac{1}{t^2}-2t>3\Leftrightarrow\frac{1-2t^3-3t^2}{t^2}>0\Leftrightarrow1-2t^3-3t^2>0\)
\(\Leftrightarrow\left(t+1\right)^2\left(1-2t\right)>0\Leftrightarrow1-2t>0\Leftrightarrow t< \frac{1}{2}\)
\(t< \frac{1}{2}\Rightarrow\sqrt{\frac{x+1}{x}}< \frac{1}{2}\Leftrightarrow\frac{x+1}{x}< \frac{1}{4}\Leftrightarrow\frac{3x+4}{4x}< 0\)
Lập bảng xét dấu ta được \(-\frac{4}{3}< x< 0\)
Kết hợp điều kiện ta được: \(-\frac{4}{3}< x< -1\) là giá trị cần tìm
3) Chứng minh BĐT phụ: \(\frac{1}{a+b}\le\frac{1}{4}\left(\frac{1}{a}+\frac{1}{b}\right)\left(a,b>0\right)\)(1)
\(\left(1\right)\Leftrightarrow\frac{1}{a+b}\le\frac{a+b}{4ab}\Leftrightarrow4ab\le\left(a+b\right)^2\Leftrightarrow\left(a-b\right)^2\ge0\)
Dấu '=' xảy ra ↔ a = b
Áp dụng BĐT trên, ta có:
\(\frac{x}{x+1}=\frac{x}{x+x+y+z}=\frac{x}{x+y+x+z}\le\frac{1}{4}\left(\frac{x}{x+y}+\frac{x}{x+z}\right)\)
Tương tự:
\(\frac{y}{y+1}\le\frac{1}{4}\left(\frac{y}{y+x}+\frac{y}{y+z}\right)\)
\(\frac{z}{z+1}\le\frac{1}{4}\left(\frac{z}{z+x}+\frac{z}{z+y}\right)\)
Cộng vế theo vế ba BĐT trên ta được:
\(P\le\frac{1}{4}\left(\frac{x}{x+y}+\frac{y}{x+y}+\frac{x}{x+z}+\frac{z}{z+x}+\frac{z}{z+y}+\frac{y}{y+z}\right)\)
\(\Leftrightarrow P\le\frac{1}{4}\left(1+1+1\right)=\frac{3}{4}\)
Dấu '=' xảy ra khi x = y = z = 1/3 (do x + y + z = 1)
Vậy GTLN của P là 3/4 khi x = y = z = 1/3
d/ Đặt \(x=a+b\) , \(y=b+c\) , \(z=c+a\)
thì : \(a=\frac{x+z-y}{2}\) ; \(b=\frac{x+y-z}{2}\) ; \(c=\frac{y+z-x}{2}\)
Ta có : \(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=\frac{\frac{x+z-y}{2}}{y}+\frac{\frac{x+y-z}{2}}{z}+\frac{\frac{y+z-x}{2}}{x}\)
\(=\frac{z+x-y}{2y}+\frac{x+y-z}{2z}+\frac{y+z-x}{2x}=\frac{1}{2}\left(\frac{x}{y}+\frac{y}{x}+\frac{z}{y}+\frac{y}{z}+\frac{z}{x}+\frac{x}{z}-3\right)\)
\(=\frac{1}{2}\left(\frac{x}{y}+\frac{y}{x}+\frac{y}{z}+\frac{z}{y}+\frac{z}{x}+\frac{x}{z}\right)-\frac{3}{2}\ge\frac{1}{2}.6-\frac{3}{2}=\frac{3}{2}\)
b/ \(a^2\left(1+b^2\right)+b^2\left(1+c^2\right)+c^2\left(1+a^2\right)\ge6abc\)
\(\Leftrightarrow\left(a^2b^2-2abc+c^2\right)+\left(b^2c^2-2abc+a^2\right)+\left(c^2a^2-2abc+b^2\right)\ge0\)
\(\Leftrightarrow\left(ab-c\right)^2+\left(bc-a\right)^2+\left(ca-b\right)^2\ge0\) (luôn đúng)
Vậy bđt ban đầu dc chứng minh.
4.
\(\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}=\frac{a^4}{ab}+\frac{b^4}{bc}+\frac{c^4}{ac}\ge\frac{\left(a^2+b^2+c^2\right)}{ab+bc+ca}\)
\(\Rightarrow\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}\ge\frac{\left(ab+bc+ca\right)^2}{ab+bc+ca}=ab+bc+ca\)
Dấu "=" xảy ra khi \(a=b=c\)
5.
\(\frac{a}{bc}+\frac{b}{ca}\ge2\sqrt{\frac{ab}{bc.ca}}=\frac{2}{c}\) ; \(\frac{a}{bc}+\frac{c}{ab}\ge\frac{2}{b}\) ; \(\frac{b}{ca}+\frac{c}{ab}\ge\frac{2}{a}\)
Cộng vế với vế:
\(2\left(\frac{a}{bc}+\frac{b}{ca}+\frac{c}{ab}\right)\ge2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
\(\Rightarrow\frac{a}{bc}+\frac{b}{ca}+\frac{c}{ab}\ge\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)
1.
Áp dụng BĐT \(x^2+y^2+z^2\ge xy+yz+zx\)
\(\Rightarrow\left(\sqrt{ab}\right)^2+\left(\sqrt{bc}\right)^2+\left(\sqrt{ca}\right)^2\ge\sqrt{ab}.\sqrt{bc}+\sqrt{ab}.\sqrt{ac}+\sqrt{bc}.\sqrt{ac}\)
\(\Rightarrow ab+bc+ca\ge\sqrt{abc}\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)\)
2.
\(\frac{ab}{c}+\frac{bc}{a}\ge2\sqrt[]{\frac{ab.bc}{ca}}=2b\) ; \(\frac{ab}{c}+\frac{ac}{b}\ge2a\) ; \(\frac{bc}{a}+\frac{ac}{b}\ge2c\)
Cộng vế với vế:
\(2\left(\frac{ab}{c}+\frac{bc}{a}+\frac{ac}{b}\right)\ge2\left(a+b+c\right)\)
\(\Leftrightarrow\frac{ab}{c}+\frac{bc}{a}+\frac{ac}{b}\ge a+b+c\)
3.
Từ câu b, thay \(c=1\) ta được:
\(ab+\frac{b}{a}+\frac{a}{b}\ge a+b+1\)
3. a) \(A=x+\frac{1}{x-1}=x-1+\frac{1}{x-1}+1\ge2\sqrt{\left(x-1\right)\cdot\frac{1}{x-1}}+1=3\)
Dấu "=" \(\Leftrightarrow x-1=\frac{1}{x-1}\Leftrightarrow x=2\)
Min \(A=3\Leftrightarrow x=2\)
b) \(B=\frac{4}{x}+\frac{1}{4y}=\frac{4}{x}+4x+\frac{1}{4y}+4y\cdot-4\left(x+y\right)\)
\(\ge2\sqrt{\frac{4}{x}\cdot4x}+2\sqrt{\frac{1}{4y}\cdot4y}-4\cdot\frac{5}{4}=5\)
Dấu "=" \(\Leftrightarrow\left\{{}\begin{matrix}\frac{4}{x}=4x\\\frac{1}{4y}=4y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=\frac{1}{4}\end{matrix}\right.\)
Min \(B=5\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=\frac{1}{4}\end{matrix}\right.\)
4. Chắc đề là tìm min???
\(C=a+b+\frac{1}{a}+\frac{1}{b}\ge a+b+\frac{4}{a+b}=a+b+\frac{1}{a+b}+\frac{3}{a+b}\)
\(\ge2\sqrt{\left(a+b\right)\cdot\frac{1}{a+b}}+\frac{3}{1}=5\)
Dấu "=" \(\Leftrightarrow\left\{{}\begin{matrix}a=b\\a+b=\frac{1}{a+b}\\a+b=1\end{matrix}\right.\Leftrightarrow a=b=\frac{1}{2}\)
Min \(C=5\Leftrightarrow a=b=\frac{1}{2}\)
1. Áp dụng bđt \(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\) ta có:
\(\left(\frac{1}{p-a}+\frac{1}{p-b}\right)+\left(\frac{1}{p-b}+\frac{1}{p-c}\right)+\left(\frac{1}{p-c}+\frac{1}{p-a}\right)\)
\(\ge\frac{4}{2p-a-b}+\frac{4}{2p-b-c}+\frac{4}{2p-a-c}\) \(=\frac{4}{c}+\frac{4}{a}+\frac{4}{b}\)
\(\Rightarrow\frac{1}{p-a}+\frac{1}{p-b}+\frac{1}{p-c}\ge2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
Dấu "=" \(\Leftrightarrow a=b=c\)
2. Áp dụng bđt Cauchy ta có :
\(a\sqrt{b-1}=a\sqrt{\left(b-1\right)\cdot1}\le a\cdot\frac{b-1+1}{2}=\frac{ab}{2}\) . Dấu "=" \(\Leftrightarrow b-1=1\Leftrightarrow b=2\)
+ Tương tự : \(b\sqrt{a-1}\le\frac{ab}{2}\). Dấu "=" \(\Leftrightarrow a=2\)
Do đó: \(a\sqrt{b-1}+b\sqrt{a-1}\le ab\). Dấu "=" \(\Leftrightarrow a=b=2\)
\(A=1-cos^2x+2cosx+1=3-\left(cosx-1\right)^2\le3\)
\(A_{max}=3\) khi \(cosx=1\)
\(B=1-sin^2x-2sin^2x-3=-1-\left(sinx+1\right)^2\le-1\)
\(B_{max}=-1\) khi \(sinx=-1\)
\(A=\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{2}\left(2cos^2\frac{x}{2}-1\right)}}}\)
\(=\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{cos^2\frac{x}{2}}}}=\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{2}cos\frac{x}{2}}}\)
\(=\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{2}\left(2cos^2\frac{x}{4}-1\right)}}\)
\(=\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{cos^2\frac{x}{4}}}=\sqrt{\frac{1}{2}+\frac{1}{2}cos\frac{x}{4}}\)
\(=\sqrt{\frac{1}{2}+\frac{1}{2}\left(2cos^2\frac{x}{8}-1\right)}=\sqrt{cos^2\frac{x}{8}}=cos\frac{x}{8}\)
\(B=\sqrt{2+\sqrt{2+\sqrt{2+2\left(2cos^2\frac{a}{2}-1\right)}}}\)
\(=\sqrt{2+\sqrt{2+\sqrt{4cos^2\frac{a}{2}}}}=\sqrt{2+\sqrt{2+2cos\frac{a}{2}}}\)
\(=\sqrt{2+\sqrt{2+2\left(cos^2\frac{a}{4}-1\right)}}=\sqrt{2+\sqrt{4cos^2\frac{a}{4}}}\)
\(=\sqrt{2+2cos\frac{a}{4}}=\sqrt{2+2\left(2cos^2\frac{a}{8}-1\right)}=2cos\frac{a}{8}\)
Bổ đề 1: Với m, n < 1 ta có bất đẳng thức:
\(\frac{1}{1+m^2}+\frac{1}{1+n^2}\le\frac{2}{1+mn}\).
Thật vậy, bất đẳng thức trên tương đương với: \(\left(mn-1\right)\left(m-n\right)^2\le0\) (luôn đúng).
Bổ đề 2: Với m, n, p < 1 ta có bất đẳng thức:
\(\frac{1}{1+m^3}+\frac{1}{1+n^3}+\frac{1}{1+p^3}\le\frac{3}{1+mnp}\left(2\right)\).
Thật vậy, áp dụng bổ đề (1) ta có:
\(VT_{\left(2\right)}=\left(\frac{1}{1+m^3}+\frac{1}{1+n^3}\right)+\left(\frac{1}{1+p^3}+\frac{1}{1+mnp}\right)-\frac{1}{1+mnp}\le\frac{2}{1+\sqrt{m^3n^3}}+\frac{2}{1+\sqrt{mnp^4}}-\frac{1}{1+mnp}\le\frac{4}{1+\sqrt[4]{m^3n^3.mnp^4}}-\frac{1}{1+mnp}=\frac{4}{1+mnp}-\frac{1}{1+mnp}=\frac{3}{1+mnp}\left(đpcm\right)\).
Quay trở lại bài toán.
Đặt \(\left(\sqrt[3]{a},\sqrt[3]{b},\sqrt[3]{c}\right)=\left(x,y,z\right)\). Ta có: \(0< x,y,z< 1\).
BĐT cần chứng minh trở thành:
\(\frac{1}{1+x^3+y^3}+\frac{1}{1+y^3+z^3}+\frac{1}{1+z^3+x^3}\le\frac{3}{1+2xyz}\left(1\right)\).
Áp dụng BĐT AM - GM và bổ đề 2 ta có: \(VT_{\left(1\right)}\le\frac{1}{1+\left(\sqrt[3]{2}\sqrt{xy}\right)^3}+\frac{1}{1+\left(\sqrt[3]{2}\sqrt{yz}\right)^3}+\frac{1}{1+\left(\sqrt[3]{2}\sqrt{zx}\right)^3}\le\frac{3}{1+\sqrt[3]{2}\sqrt[3]{2}\sqrt[3]{2}\sqrt{xy.yz.zx}}=\frac{3}{1+2xyz}=VP_{\left(1\right)}\left(đpcm\right)\)
Bạn bổ sung cho mình thêm điều kiện ở hai bổ đề:
Bổ đề 1: Thêm m, n > 0.
Bổ đề 2: Thêm m, n, p > 0.