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Bài 1:
Ta có: a + b - 2c = 0
⇒ a = 2c − b thay vào a2 + b2 + ab - 3c2 = 0 ta có:
(2c − b)2 + b2 + (2c − b).b − 3c2 = 0
⇔ 4c2 − 4bc + b2 + b2 + 2bc − b2 − 3c2 = 0
⇔ b2 − 2bc + c2 = 0
⇔ (b − c)2 = 0
⇔ b − c = 0
⇔ b = c
⇒ a + c − 2c = 0
⇔ a − c = 0
⇔ a = c
⇒ a = b = c
Vậy a = b = c
\(\frac{a+bc}{b+c}+\frac{b+ac}{c+a}+\frac{c+ab}{a+b}\)
\(=\frac{a\left(a+b+c\right)+bc}{b+c}+\frac{b\left(a+b+c\right)+ac}{a+c}+\frac{c\left(a+b+c\right)+ab}{a+b}\)
\(=\frac{\left(a+b\right)\left(a+c\right)}{b+c}+\frac{\left(a+b\right)\left(b+c\right)}{a+c}+\frac{\left(c+a\right)\left(c+b\right)}{a+b}\)
Áp dụng bđt Cô Si: \(\frac{\left(a+b\right)\left(a+c\right)}{b+c}+\frac{\left(a+b\right)\left(b+c\right)}{a+c}\ge2\left(a+b\right)\)
Tương tự,cộng theo vế và rút gọn =>đpcm
\(\frac{a+bc}{b+c}+\frac{b+ac}{c+a}+\frac{c+ab}{a+b}\)
\(=\frac{a\left(a+b+c\right)+bc}{b+c}+\frac{b\left(a+b+c\right)+ac}{a+c}+\frac{c\left(a+b+c\right)+ab}{a+b}\)
\(=\frac{\left(a+b\right)\left(a+c\right)}{b+c}+\frac{\left(a+b\right)\left(b+c\right)}{a+c}+\frac{\left(c+a\right)\left(c+b\right)}{a+b}\)
Áp dụng bđt CÔ si
\(\frac{\left(a+b\right)\left(a+c\right)}{b+c}+\frac{\left(a+b\right)\left(b+c\right)}{a+c}\ge2\left(a+b\right)\)
.............
\(a^3+1+1\ge3a\)
\(b^3+1+1\ge3b\)
\(c^3+1+1\ge3c\)
\(2\left(a^3+b^3+c^3\right)\ge6abc\)
Cộng vế:
\(3\left(a^3+b^3+c^3\right)+6\ge3\left(a+b+c+2abc\right)=15\)
\(\Rightarrow a^3+b^3+c^3\ge3\) (đpcm)
Dấu "=" xảy ra khi \(a=b=c=1\)
\(A=\left(1+b^2+a^2+a^2b^2\right).\left(1+c^2\right)\)
\(=1+a^2+b^2+c^2+a^2c^2+b^2c^2+a^2b^2+a^2b^2c^2\)
\(=1+\left(a+b+c\right)^2-2.\left(ab+bc+ac\right)+\left(ab+bc+ac\right)^2-2abc.\left(a+b+c\right)+a^2b^2c^2\)
Thay ab+bc+ac=1 vào A, ta có:
\(A=1+\left(a+b+c\right)^2-2+1-2abc.\left(a+b+c\right)+a^2b^2c^2\)
\(=\left(a+b+c\right)^2-2abc.\left(a+b+c\right)+a^2b^2c^2\)
\(=\left(a+b+c-abc\right)^2\)
Vì a,b,c thuộc Z
\(\Rightarrow\left(a+b+c-abc\right)^2\)là số chính phương
\(\hept{\begin{cases}\left(1+a^2\right)=\left(ab+bc+ca+a^2\right)=b\left(a+c\right)+a\left(a+c\right)=\left(a+b\right)\left(a+c\right)\\\left(1+b^2\right)=\left(ab+bc+ca+b^2\right)=a\left(b+c\right)+b\left(b+c\right)=\left(a+b\right)\left(b+c\right)\\\left(1+c^2\right)=\left(ab+bc+ca+c^2\right)=a\left(b+c\right)+c\left(b+c\right)=\left(a+c\right)\left(b+c\right)\end{cases}}\)
\(\Rightarrow A=\text{[}\left(a+b\right)\left(b+c\right)\left(c+a\right)\text{]}^2\Rightarrow\text{đ}pcm\)
Áp dụng BĐT Cauchy-Schwarz dạng Engel ta có:
\(\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}\ge\frac{\left(1+1+1\right)^2}{3+a+b+c+}=\frac{9}{6}=\frac{3}{2}\)
Bạn xem lại đề nhé :
Phương trình \(b^3-3b^2+5b+11=0\)không có nghiệm dương nhé
\(VT=b\left(b-\frac{3}{2}\right)^2+\frac{11}{4}b+11>0\forall b>0\)
\(xy+x+1=3y\Rightarrow x+\dfrac{1}{y}+\dfrac{x}{y}=3\)
Ta có:
\(x^3+1+1\ge3x\)
\(\dfrac{1}{y^3}+1+1\ge\dfrac{3}{y}\)
\(x^3+\dfrac{1}{y^3}+1\ge\dfrac{3x}{y}\)
Cộng vế:
\(2\left(x^3+\dfrac{1}{y^3}\right)+5\ge3\left(x+\dfrac{1}{y}+\dfrac{x}{y}\right)=9\)
\(\Rightarrow x^3+\dfrac{1}{y^3}\ge2\)
\(\Rightarrow x^3y^3+1\ge2y^3\) (đpcm)
Dấu "=" xảy ra khi \(x=y=1\)
\(gt\Rightarrow\left(a+b\right)^2=1\Leftrightarrow a^2+2ab+b^2=1\) (1)
Do theo BĐT AM-GM (Cô si) \(a^2+b^2\ge2\left|ab\right|\ge2ab\)
Thay vào (1) suy ra \(1=a^2+2ab+b^2\ge4ab\)
Suy ra \(ab\le\frac{1}{4}\).Từ đây ta có: \(a^2+b^2=\left(a+b\right)^2-2ab=1-2ab\ge\frac{1}{2}^{\left(đpcm\right)}\)
Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}a^2=b^2\\a+b=1\end{cases}}\Leftrightarrow\hept{\begin{cases}a=b\\a+b=1\end{cases}}\Leftrightarrow a=b=\frac{1}{2}\)
Phép chứng minh hoàn tất!