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5. phân tích ra : \(1+\dfrac{a}{b}+\dfrac{b}{a}+1\)
áp dụng bđ cosy
\(\dfrac{a}{b}+\dfrac{b}{a}\ge2\sqrt{\dfrac{a}{b}.\dfrac{b}{a}}=2\)
=> đpcm
6. \(x^2-x+1=x^2-2.\dfrac{1}{2}.x+\dfrac{1}{4}+\dfrac{3}{4}=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\)
hay với mọi x thuộc R đều là nghiệm của bpt
7.áp dụng bđt cosy
\(a^4+b^4+c^4+d^4\ge2\sqrt{a^2.b^2.c^2.d^2}=4abcd\left(đpcm\right)\)
\(\Leftrightarrow a\left(-\dfrac{8}{a^2+16}+\dfrac{a+8}{2a^2+16}\right)>=0\)
=>\(\dfrac{a^2\left(a-4\right)^2}{\left(a^2+16\right)\left(2a^2+16\right)}>=0\)(luôn đúng)
Lời giải:
Áp dụng BĐT AM-GM:
$1=x+y\geq 2\sqrt{xy}\Rightarrow xy\leq \frac{1}{4}$
$P=x^2y^2+\frac{1}{x^2y^2}+2-\frac{17}{6}$
$=x^2y^2+\frac{1}{x^2y^2}-\frac{5}{6}$
$=(x^2y^2+\frac{1}{256x^2y^2})+\frac{255}{256x^2y^2}-\frac{5}{6}$
$\geq 2\sqrt{\frac{1}{256}}+\frac{255}{256.\frac{1}{4^2}}-\frac{5}{6}=\frac{731}{48}$
Vậy $P_{\min}=\frac{731}{48}$ khi $x=y=\frac{1}{2}$
2.
Từ giả thiết, ta có :
\(\frac{1}{1+a}\ge1-\frac{1}{1+b}+1-\frac{1}{1+c}+1-\frac{1}{1+d}\)
\(=\frac{b}{1+b}+\frac{c}{1+c}+\frac{d}{1+d}\ge3\sqrt[3]{\frac{b.c.d}{\left(1+b\right)\left(1+c\right)\left(1+d\right)}}\)
Tương tự, ta cũng có :
\(\frac{1}{1+b}\ge3\sqrt[3]{\frac{c.d.a}{\left(1+c\right)\left(1+d\right)\left(1+a\right)}}\)
\(\frac{1}{1+c}\ge3\sqrt[3]{\frac{abd}{\left(1+a\right)\left(1+b\right)\left(1+d\right)}}\)
\(\frac{1}{1+d}\ge3\sqrt[3]{\frac{abc}{\left(1+a\right)\left(1+b\right)\left(1+c\right)}}\)
Nhân vế theo vế 4 BĐT vừa chững minh rồi rút gọn ta được :
\(abcd\le\frac{1}{81}\left(đpcm\right)\)
2) Từ \(\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}+\frac{1}{1+d}\ge3.\)
\(\Rightarrow\frac{1}{1+a}\ge\left(1-\frac{1}{1+b}\right)+\left(1-\frac{1}{1+c}\right)+\left(1-\frac{1}{1+d}\right)\)
\(=\frac{b}{1+b}+\frac{c}{1+c}+\frac{d}{1+d}\ge3\sqrt[3]{\frac{bcd}{\left(1+b\right)\left(1+c\right)\left(1+d\right)}}.\)(BĐT AM-GM)
Tương tự :
\(\frac{1}{1+b}\ge3\sqrt[3]{\frac{acd}{\left(1+a\right)\left(1+c\right)\left(1+d\right)}}\)
\(\frac{1}{1+c}\ge3\sqrt[3]{\frac{abd}{\left(1+a\right)\left(1+b\right)\left(1+d\right)}}\)
\(\frac{1}{1+d}\ge3\sqrt[3]{\frac{abc}{\left(1+a\right)\left(1+b\right)\left(1+c\right)}}.\)
Từ đó suy ra:
\(\frac{1}{1+a}.\frac{1}{1+b}.\frac{1}{1+c}.\frac{1}{1+d}\ge3.3.3.3\sqrt[3]{\frac{\left(abcd\right)^3}{\left[\left(1+a\right)\left(1+b\right)\left(1+c\right)\left(1+d\right)\right]^3}}\)
\(\Leftrightarrow\frac{1}{\left(1+a\right)\left(1+b\right)\left(1+c\right)\left(1+d\right)}\ge\frac{81abcd}{\left(1+a\right)\left(1+b\right)\left(1+c\right)\left(1+d\right)}.\)
\(\Leftrightarrow81abcd\le1\Leftrightarrow abcd\le\frac{1}{81}\)
Dấu '=' xảy ra khi \(a=b=c=d=\frac{1}{3}.\)
3)Ta có: \(\left(\sqrt{a}+\sqrt{b}\right)^8=\left[\left(\sqrt{a}+\sqrt{b}\right)^2\right]^4=\left(a+b+2\sqrt{ab}\right)^4.\)(1)
Với \(a,b\ge0\),áp dụng BĐT AM-GM cho (a+b) và (\(2\sqrt{ab}\)) ta được
\(\left(a+b\right)+2\sqrt{ab}\ge2\sqrt{\left(a+b\right)2\sqrt{ab}}\)(2)
Từ (1) và (2) suy ra:
\(\left(\sqrt{a}+\sqrt{b}\right)^8\ge\left(2\sqrt{\left(a+b\right)2\sqrt{ab}}\right)^4\)
\(\Leftrightarrow\left(\sqrt{a}+\sqrt{b}\right)^8\ge64ab\left(a+b\right)^2.\)
Dấu '=' xảy ra khi \(a+b=2\sqrt{ab}\Leftrightarrow a=b\)
1) Với \(x\le\frac{2}{3}\Rightarrow2-3x\ge0\)
Khi đó ,áp dụng bất đẳng thức AM-GM cho 2 số ta được:
\(\left(2-3x\right)+\frac{9}{2-3x}\ge2\sqrt{\left(2-3x\right)\frac{9}{2-3x}}=2.3=6\)
\(\Leftrightarrow2+\left(2-3x\right)+\frac{9}{2-3x}\ge2+6\)
\(\Leftrightarrow4-3x+\frac{9}{2-3x}\ge8\)
Dấu '=' xảy ra khi \(2-3x=\frac{9}{2-3x}\Leftrightarrow\left(2-3x\right)^2=9\Leftrightarrow2-3x=3\Leftrightarrow x=-\frac{1}{3}\)( vì 2-3x>0)
Câu 1:
Ta có: \(\left(\dfrac{a+b}{2}\right)^2\ge ab\)
\(\Leftrightarrow\dfrac{\left(a+b\right)^2}{2^2}-ab\ge0\)
\(\Leftrightarrow\dfrac{a^2+2ab+b^2-4ab}{4}\ge0\)
\(\Leftrightarrow\dfrac{a^2-2ab+b^2}{4}\ge0\)
\(\Leftrightarrow\dfrac{\left(a-b\right)^2}{4}\ge0\)
Vì \(\left(a-b\right)^2\ge0\forall a,b\)
\(\Rightarrow\dfrac{\left(a-b\right)^2}{4}\ge0\forall a,b\)
\(\Rightarrow\left(\dfrac{a+b}{2}\right)^2\ge ab\) (1)
Ta có: \(\dfrac{a^2+b^2}{2}\ge\left(\dfrac{a+b}{2}\right)^2\)
\(\Leftrightarrow\dfrac{a^2+b^2}{2}-\dfrac{\left(a+b\right)^2}{4}\ge0\)
\(\Leftrightarrow\dfrac{2a^2-2b^2-a^2-2ab-b^2}{4}\ge0\)
\(\Leftrightarrow\dfrac{a^2-2ab-b^2}{4}\ge0\)
\(\Leftrightarrow\dfrac{\left(a-b\right)^2}{4}\ge0\)
Vì \(\left(a-b\right)^2\ge0\forall a,b\)
\(\Rightarrow\dfrac{\left(a-b\right)^2}{4}\ge0\forall a,b\)
\(\Rightarrow\dfrac{a^2+b^2}{2}\ge\left(\dfrac{a+b}{2}\right)^2\) (2)
Từ (1) và (2) \(\Rightarrow ab\le\left(\dfrac{a+b}{2}\right)^2\le\dfrac{a^2+b^2}{2}\)
5 , a3+b3+c3\(\ge\) 3abc
\(\Leftrightarrow\) a3+3a2b+3ab2+b3+c3-3a2b-3ab2-3abc\(\ge\) 0
\(\Leftrightarrow\) (a+b)3+c3-3ab(a+b+c) \(\ge0\)
\(\Leftrightarrow\) (a+b+c)(a2+2ab+b2-ac-bc+c2)-3ab(a+b+c) \(\ge0\)
\(\Leftrightarrow\) (a+b+c)(a2+b2+c2-ab-bc-ca)\(\ge0\) (1)
ta co : a,b,c>0 \(\Rightarrow\)a+b+c>0 (2)
(a-b)2+(b-c)2+(c-a)2\(\ge0\)
<=> 2a2+2b2+2c2-2ac-2cb-2ab\(\ge0\)
<=>a2+b2+c2-ab-bc-ac\(\ge\) 0 (3)
Từ (1)(2)(3)=> pt luôn đúng
Áp dụng bđt Cauchy Schwarz dạng Engel ta có:
\(\left(a+b+c\right)\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{a+c}\right)\ge\left(a+b+c\right).\dfrac{\left(1+1+1\right)^2}{2\left(a+b+c\right)}\)
\(\ge\dfrac{9}{2}\left(đpcm\right)\)
Sử dụng AM-GM, ta có:
\(\left(a+\dfrac{1}{b}\right)\left(b+\dfrac{1}{a}\right)\ge4\Rightarrow b+\dfrac{1}{a}\ge4\)
Sử dụng Cauchy-Schwarz, ta có:
\(A\ge\dfrac{\left(a+\dfrac{1}{a}+b+\dfrac{1}{b}\right)^2}{2}\ge\dfrac{\left(1+4\right)^2}{2}=\dfrac{25}{2}\)
Đẳng thức xảy ra khi \(a=\dfrac{1}{2};b=2\)
=>(ab-1)^2+ab(a-b)^2>=0
=>a^2b^2-2ab+1+ab(a^2-2ab+b^2)>=0
=>a^2b^2-2ab+1+a^3b-2a^2b^2+ab^3>=0
=>a^3b+ab^3-a^2b^2-2ab+1>=0
=>ab(a^2+b^2)-2ab-a^2b^2+1>=0
=>ab(a^2+b^2-2-ab)+1>=0(luôn đúng)