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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)
Bài 1:
Áp dụng BĐT AM-GM ta có:
\(1=\left(a+b+c\right)^2=\left[a+\left(b+c\right)\right]^2\ge4a\left(b+c\right)\)
\(\Leftrightarrow b+c\ge4a\left(b+c\right)^2\). Và \(\left(b+c\right)^2\ge4bc\)
\(\Rightarrow b+c\ge4a\left(b+c\right)^2\ge4a\cdot4bc=16abc\)
Bài 2:
Áp dụng BĐT Cauchy-Schwarz dạng Engel ta có:
\(VT=\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{4}{c}+\dfrac{16}{d}=\dfrac{1^2}{a}+\dfrac{1^2}{b}+\dfrac{2^2}{c}+\dfrac{4^2}{d}\)
\(\ge\dfrac{\left(1+1+2+4\right)^2}{a+b+c+d}=\dfrac{8^2}{a+b+c+d}=64=VP\)
Bài 1 :Áp dụng Bất Đẳng Thức (x+y)² ≥ 4xy cho hai số không âm có
1 = (a + b+ c)² ≥ 4a(b + c)
--> b + c ≥ 4a(b + c)²
Mà (b + c)² ≥ 4bc
Vậy b + c ≥ 16abc.
Bài 2 bạn Ace Legona làm ròi mình ko làm lại
Chúc bạn học tốt 
a) Áp dụng BĐT bunhiacopxki ta có:
A= \(a^2+b^2\) \(\geq\) \(\dfrac{\left(a+b\right)^2}{2}=\dfrac{1}{2}\)
Vậy Min A= \(\dfrac{1}{2}\) khi a=b=\(\dfrac{1}{2}\)
b) Ta có: B= \(\dfrac{a^2}{b}+\dfrac{b^2}{a}\)
\(\Leftrightarrow\) B= \(\left(\dfrac{a^2}{b}+b\right)+\left(\dfrac{b^2}{a}+a\right)-\left(a+b\right)\) \(\geq\) \(2\sqrt{\dfrac{a^2}{b}.b}+2\sqrt{\dfrac{b^2}{a}.a}-a-b\) = \(2a+2b-a-b\) \(=a+b=1\)
Từ đó suy ra: \(\dfrac{a^2}{b}+\dfrac{b^2}{a}\) \(\geq\) 1
Vậy Min B = 1 khi a=b=\(\dfrac{1}{2}\)
\(\dfrac{a^2+b^2+c^2}{3}\ge\left(\dfrac{a+b+c}{3}\right)^2\)
\(\Leftrightarrow3\left(a^2+b^2+c^2\right)\ge\left(a+b+c\right)^2\)
\(\Leftrightarrow2\left(a^2+b^2+c^2\right)-2\left(ab+bc+ca\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\) (đúng)
\(\dfrac{a^2+b^2+c^2}{3}\ge\left(\dfrac{a+b+c}{3}^{ }\right)^2\\ \Leftrightarrow\dfrac{a^2+b^2+c^2}{3}\ge\dfrac{a^2+b^2+c^2+2ab+2bc+2ac}{9}\\ \Leftrightarrow\dfrac{3a^2+3b^2+3c^2-a^2-b^2-c^2-2ab-2ac-2bc}{9}\ge0\\ \Leftrightarrow\dfrac{2a^2+2b^2+2c^2-2ab-2ac-2bc}{9}\ge0\\ \Leftrightarrow\dfrac{2\left(a^2+b^2+c^2\right)-2\left(ab+ac+bc\right)}{9}\ge0\)
mà ta có:
\(a^2+b^2+c^2\ge ab+bc+ac\\ \Rightarrow2\left(a^2+b^2+c^2\right)-2\left(ab+bc+ac\right)\ge0\\ \Rightarrow\dfrac{2\left(a^2+b^2+c^2\right)}{9}\ge0\)
Vậy \(\dfrac{a^2+b^2+c^2}{3}\ge\left(\dfrac{a+b+c}{3}\right)^2\)
a. Xét hiệu: \(\dfrac{1}{a}+\dfrac{1}{b}-\dfrac{4}{a+b}\)
=\(\dfrac{b\left(a+b\right)+a\left(a+b\right)-4ab}{ab\left(a+b\right)}\)
\(=\dfrac{a^2-2ab+b^2}{ab\left(a+b\right)}=\dfrac{\left(a-b\right)^2}{ab\left(a+b\right)}\ge0\)
Vì a,b>0
Xảy ra đẳng thức khi và chỉ khi a=b
a) Ta có: \(\left(a-b\right)^2\ge0\left(1\right)\forall a,b\)
( Dấu = xày ra khi và chỉ khi a=b)
Cộng 4ab vào 2 vế, ta có:
\(\left(a-b\right)^2+4ab\ge4ab\Leftrightarrow\left(a+b\right)^2\ge4ab\)
Chia 2 vế cho ab(a+b)>0, ta có:
\(\dfrac{a+b}{ab}\ge\dfrac{4}{a+b}\Leftrightarrow\)\(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\)
b) Ta có:
\(2p=a+b+c\)
\(p-a=\dfrac{a+b+c}{2}-a=\dfrac{b+c-a}{2}>0\) vì b+c>a
Tương tự: \(p-b>0,p-c>0\)
Áp dụng BĐT: \(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\)cho từng cặp số p-a, p-b; p-b,p-c;p-c,p-a
Ta có:
\(\dfrac{1}{p-a}+\dfrac{1}{p-b}\ge\dfrac{4}{\left(p-a\right)+\left(p-b\right)}=\dfrac{4}{2p-\left(a+b\right)}=\dfrac{4}{c}\left(1\right)\)
Tương tự:
\(\dfrac{1}{p-b}+\dfrac{1}{p-c}\ge\dfrac{4}{a}\left(2\right)\)
\(\dfrac{1}{p-c}+\dfrac{1}{p-a}\ge\dfrac{4}{b}\left(3\right)\)
Cộng các BĐT cùng chiều (1), (2), (3) vế theo vế, ta có:
\(2\left(\dfrac{1}{p-a}+\dfrac{1}{p-b}+\dfrac{1}{p-c}\right)\ge4\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
Do đó: \(\dfrac{1}{p-a}+\dfrac{1}{p-b}+\dfrac{1}{p-c}\ge2\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
Từ \(p=\dfrac{a+b+c}{2}\Rightarrow2p=a+b+c\)
Áp dụng BĐT \(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{\left(1+1\right)^2}{a+b}=\dfrac{4}{a+b}\) ta có:
\(\dfrac{1}{p-a}+\dfrac{1}{p-b}\ge\dfrac{4}{p-a+p-b}=\dfrac{4}{2q-a-b}\)
\(=\dfrac{4}{a+b+c-a-b}=\dfrac{4}{c}\). Tương tự cho 2 BĐT còn lại:
\(\dfrac{1}{p-b}+\dfrac{1}{p-c}\ge\dfrac{4}{a};\dfrac{1}{p-c}+\dfrac{1}{p-a}\ge\dfrac{4}{b}\)
Cộng theo vế 3 BĐT trên ta có:
\(2\left(\dfrac{1}{p-a}+\dfrac{1}{p-b}+\dfrac{1}{p-c}\right)\ge4\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
\(\Leftrightarrow\dfrac{1}{p-a}+\dfrac{1}{p-b}+\dfrac{1}{p-c}\ge2\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
Đẳng thức xảy ra khi \(a=b=c\)