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ta có A=\(\frac{1}{a^2+2a+2+b^2}+\frac{1}{b^2+2b+2+c^2}+\frac{1}{c^2+2c+2+a^2}\)
Áp dụng bđt cô si, ta có \(a^2+b^2\ge2ab\) =>\(\frac{1}{a^2+b^2+2a+2}\le\frac{1}{2ab+2a+2}\)
tương tự, rồi + vào, ta có
A \(\le\frac{1}{2}\left(\frac{1}{a+ab+1}+\frac{1}{b+bc+1}+\frac{1}{c+ca+1}\right)\)
mà với abc=1 thì ta luôn chứng minh được \(\frac{1}{a+ab+1}+\frac{1}{b+bc+1}+\frac{1}{c+ca+1}=1\)
=> A <= 1/2 (ĐPCM)
dấu = xảy ra <=> a=b=c=1
^_^
Áp dụng BĐT Cauchy-Schwarz ta có:
\(\left(a+b+c\right)\left(a+a^2b+\frac{1}{c}\right)\ge\left(ab+a+1\right)^2\)
Mà \(\left(a+b+c\right)\left(a+a^2b+\frac{1}{c}\right)=\left(a+b+c\right)\left(a+a^2b+ab\right)\)
\(\Rightarrow\frac{a}{\left(ab+a+1\right)^2}\ge\frac{a}{\left(a+b+c\right)\left(a+a^2b+ab\right)}=\frac{1}{\left(a+b+c\right)\left(1+ab+b\right)}\)
Tương tự rồi cộng theo vế 3 BĐT ta có:
\(VT\ge\frac{1}{a+b+c}\left(Σ\frac{1}{1+ab+b}\right)=\frac{1}{a+b+c}\left(abc=1\right)\)
Đẳng thức xảy ra khi \(a=b=c=1\)
nhầm lẫn 1 số chỗ nên giờ mới ra,mong bn thông cảm
ta có:
\(\frac{a}{ab+a+1}+\frac{b}{bc+b+1}+\frac{c}{ca+c+1}=\frac{1}{bc+b+1}+\frac{b}{bc+b+1}+\frac{bc}{bc+b+1}=1\)
đặt \(P=\frac{a}{\left(ab+a+1\right)^2}+\frac{b}{\left(bc+b+1\right)^2}+\frac{c}{\left(ca+c+1\right)^2}\)
áp dụng bunhia ta có:
\(P\left(a+b+c\right)\ge\left(\frac{a}{ab+a+1}+\frac{b}{bc+b+1}+\frac{c}{ca+c+1}\right)^2=1\)
\(\Rightarrow P\ge\frac{1}{a+b+c}\)
Áp dụng BĐT AM-GM ta có:
\(\frac{1}{a^2\left(b+c\right)}+\frac{b+c}{4}\ge2\sqrt{\frac{1}{a^2\left(b+c\right)}\cdot\frac{b+c}{4}}=2\cdot\frac{1}{2a}=\frac{1}{a}\)
Tuong tu cho 2 BDT con lai ta cung co
\(\frac{1}{b^2\left(a+c\right)}+\frac{a+c}{4}\ge\frac{1}{b};\frac{1}{c^2\left(a+b\right)}+\frac{a+b}{4}\ge\frac{1}{c}\)
Cong theo ve cac BDT tren ta co
\(VT+\frac{2\left(a+b+c\right)}{4}\ge\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)
\(\Rightarrow VT+\frac{a+b+c}{2}\ge3\sqrt[3]{\frac{1}{abc}}=3\left(abc=1\right)\)
\(\Rightarrow VT+\frac{3\sqrt[3]{abc}}{2}\ge3\Rightarrow VT+\frac{3}{2}\ge3\Rightarrow VT\ge\frac{3}{2}\)
Dang thuc xay ra khi \(a=b=c=1\)
Áp dụng BĐT Cô-si cho 3 số dương, ta có :
\(\frac{1}{a\left(a+b\right)}+\frac{1}{b\left(b+c\right)}+\frac{1}{c\left(a+c\right)}\ge3\sqrt[3]{\frac{1}{abc\left(a+b\right)\left(b+c\right)\left(a+c\right)}}\)
Cần chứng minh : \(\sqrt[3]{\frac{1}{abc\left(a+b\right)\left(b+c\right)\left(a+c\right)}}\ge\frac{9}{2\left(a+b+c\right)^2}\)
hay \(8\left(a+b+c\right)^6\ge729abc\left(a+b\right)\left(b+c\right)\left(a+c\right)\)
Thật vậy, ta có : \(\left(a+b+c\right)^3\ge\left(3\sqrt[3]{abc}\right)^3=27abc\)
\(8\left(a+b+c\right)^3=\left(2\left(a+b+c\right)\right)^3=\left(a+b+b+c+a+c\right)^3\)
\(\ge\left(3\sqrt[3]{\left(a+b\right)\left(b+c\right)\left(a+c\right)}\right)^3=27\left(a+b\right)\left(b+c\right)\left(a+c\right)\)
Nhân từng vế 2 bất đẳng thức trên, ta được đpcm
Dấu "=" xảy ra khi a = b = c
Vậy ...
2. Áp dụng BĐT Cô-si cho 3 số không âm, ta có :
\(B\ge3\sqrt[3]{\sqrt{\left(a^3+b^3+1\right)\left(b^3+c^3+1\right)\left(a^3+c^3+1\right)}}\)
Ta có : \(a^3+b^3+1\ge3\sqrt[3]{a^3b^3}=3ab\Rightarrow\sqrt{a^3+b^3+1}\ge\sqrt{3ab}\)
Tương tự : ....
\(\Rightarrow\sqrt{\left(a^3+b^3+1\right)\left(b^3+c^3+1\right)\left(c^3+a^3+1\right)}\ge\sqrt{27a^2b^2c^2}=\sqrt{27}\)
\(\Rightarrow B\ge3\sqrt[3]{\sqrt{27}}=3\sqrt{3}\)
Vậy GTNN của B là \(3\sqrt{3}\)khi a = b = c = 1
Bài 2:b) \(9=\left(\frac{1}{a^3}+1+1\right)+\left(\frac{1}{b^3}+1+1\right)+\left(\frac{1}{c^3}+1+1\right)\)
\(\ge3\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\therefore\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\le3\)
Ta sẽ chứng minh \(P\le\frac{1}{48}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2\)
Ai có cách hay?
1/Đặt a=1/x,b=1/y,c=1/z ->x+y+z=1.
2a) \(VT=\frac{\left(\frac{1}{a^3}+\frac{1}{b^3}\right)\left(\frac{1}{a}+\frac{1}{b}\right)}{\frac{1}{a}+\frac{1}{b}}\ge\frac{\left(\frac{1}{a^2}+\frac{1}{b^2}\right)^2}{\frac{1}{a}+\frac{1}{b}}\)
\(=\frac{\left[\frac{\left(a^2+b^2\right)^2}{a^4b^4}\right]}{\frac{a+b}{ab}}=\frac{\left(a^2+b^2\right)^2}{a^3b^3\left(a+b\right)}\ge\frac{\left(a+b\right)^3}{4\left(ab\right)^3}\)
\(\ge\frac{\left(a+b\right)^3}{4\left[\frac{\left(a+b\right)^2}{4}\right]^3}=\frac{16}{\left(a+b\right)^3}\)
Tìm GTLN ko phải tìm GTNN
Ta có: \(\frac{1}{ab+a+1}+\frac{1}{bc+b+1}+\frac{1}{ca+c+1}=1\) (*)
Lại có: \(\left(a+1\right)^2+b^2+1=a^2+b^2+2a+2\ge2ab+2a+2=2\left(ab+a+1\right)\)
\(\Rightarrow\frac{1}{\left(a+1\right)^2+b^2+1}\le\frac{1}{2\left(ab+a+1\right)}\) tương tự ta có:
\(\frac{1}{\left(b+1\right)^2+c^2+1}\le\frac{1}{2\left(bc+b+1\right)};\frac{1}{\left(c+1\right)^2+a^2+1}\le\frac{1}{2\left(ca+c+1\right)}\)
Cộng theo vế ta có: \(P\le\frac{1}{2\left(ab+a+1\right)}+\frac{1}{2\left(bc+b+1\right)}+\frac{1}{2\left(ca+c+1\right)}\)
\(=\frac{1}{2}\left(\frac{1}{ab+a+1}+\frac{1}{bc+b+1}+\frac{1}{ca+c+1}\right)=\frac{1}{2}\) theo (*)
Dấu "=" khi a=b=c=1