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Từ đề bài ta có:
\(2\sqrt{xy}\le x+y=1\)
\(\Rightarrow xy\le\dfrac{1}{4}\)
Ta có:
\(P=\left(1-\dfrac{1}{x^2}\right)\left(1-\dfrac{1}{y^2}\right)=\dfrac{1-x^2-y^2+x^2y^2}{x^2y^2}\)
\(=1+\dfrac{-\left(x+y\right)^2+2xy+1}{x^2y^2}\)
\(=1+\dfrac{2}{xy}\ge1+8=9\)
Vậy GTNN là A = 9 khi \(x=y=\dfrac{1}{2}\)
Áp dụng BĐT AM-GM ta có:
\(\dfrac{x+1}{1+y^2}=x+1-\dfrac{y^2\left(x+1\right)}{y^2+1}\ge x+1-\dfrac{y\left(x+1\right)}{2}\)
Tương tự cho 2 BĐT còn lại ta cũng có:
\(\dfrac{y+1}{1+z^2}\ge y+1-\dfrac{z\left(y+1\right)}{2};\dfrac{z+1}{1+x^2}\ge z+1-\dfrac{x\left(z+1\right)}{2}\)
Cộng theo vế 3 BĐT trên ta có:
\(Q\ge\left(x+y+z+3\right)-\dfrac{x\left(z+1\right)+y\left(x+1\right)+z\left(y+1\right)}{2}\)
\(=6-\dfrac{xy+yz+xz+x+y+z}{2}\)
\(\ge6-\dfrac{\dfrac{\left(x+y+z\right)^2}{3}+3}{2}=6-3=3\)
Đẳng thức xảy ra khi \(x=y=z=1\)
\(A=\dfrac{1}{x^2+y^2}+\dfrac{1}{2xy}+\dfrac{1}{2xy}\ge\dfrac{4}{x^2+2xy+y^2}+\dfrac{1}{2.\dfrac{\left(x+y\right)^2}{4}}=4+2=6\)
12. Ta có \(ab\le\frac{a^2+b^2}{2}\)
=> \(a^2-ab+3b^2+1\ge\frac{a^2}{2}+\frac{5}{2}b^2+1\)
Lại có \(\left(\frac{a^2}{2}+\frac{5}{2}b^2+1\right)\left(\frac{1}{2}+\frac{5}{2}+1\right)\ge\left(\frac{a}{2}+\frac{5}{2}b+1\right)^2\)
=> \(\sqrt{a^2-ab+3b^2+1}\ge\frac{a}{4}+\frac{5b}{4}+\frac{1}{2}\)
=> \(\frac{1}{\sqrt{a^2-ab+3b^2+1}}\le\frac{4}{a+b+b+b+b+b+1+1}\le\frac{4}{64}.\left(\frac{1}{a}+\frac{5}{b}+2\right)\)
Khi đó
\(P\le\frac{1}{16}\left(6\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)+6\right)\le\frac{3}{2}\)
Dấu bằng xảy ra khi a=b=c=1
Vậy \(MaxP=\frac{3}{2}\)khi a=b=c=1
13. Ta có \(\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\le1\)
\(\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\ge\frac{9}{a+b+c+3}\)( BĐT cosi)
=> \(1\ge\frac{9}{a+b+c+3}\)
=> \(a+b+c\ge6\)
Ta có \(a^3-b^3=\left(a-b\right)\left(a^2+ab+b^2\right)\)
=> \(\frac{a^3-b^3}{a^2+ab+b^2}=a-b\)
Tương tự \(\frac{b^3-c^3}{b^2+bc+c^2}=b-c\),,\(\frac{c^3-a^2}{c^2+ac+a^2}=c-a\)
Cộng 3 BT trên ta có
\(\frac{a^3}{a^2+ab+b^2}+\frac{b^3}{b^2+bc+c^2}+\frac{c^3}{c^2+ac+c^2}=\frac{b^3}{a^2+ab+b^2}+\frac{c^3}{c^2+bc+b^2}+\frac{a^3}{a^2+ac+c^2}\)
Khi đó \(2P=\frac{a^3+b^3}{a^2+ab+b^2}+...\)
=> \(2P=\frac{\left(a+b\right)\left(a^2-ab+b^2\right)}{a^2+ab+b^2}+....\)
Xét \(\frac{a^2-ab+b^2}{a^2+ab+b^2}\ge\frac{1}{3}\)
<=> \(3\left(a^2-ab+b^2\right)\ge a^2+ab+b^2\)
<=> \(a^2+b^2\ge2ab\)(luôn đúng )
=> \(2P\ge\frac{1}{3}\left(a+b+b+c+a+c\right)=\frac{2}{3}.\left(a+b+c\right)\ge4\)
=> \(P\ge2\)
Vậy \(MinP=2\)khi a=b=c=2
Lưu ý : Chỗ .... là tương tự
\(B=\left(1-\dfrac{1}{x^2}\right)\left(1-\dfrac{1}{y^2}\right)\)
\(=\left(1-\dfrac{1}{x}\right)\left(1+\dfrac{1}{x}\right)\left(1-\dfrac{1}{y}\right)\left(1+\dfrac{1}{y}\right)\)
\(=\dfrac{x-1}{x}\cdot\dfrac{y-1}{y}\cdot\left(1+\dfrac{1}{x}\right)\left(1+\dfrac{1}{y}\right)\)
\(=\dfrac{x-\left(x+y\right)}{x}\cdot\dfrac{y-\left(x+y\right)}{y}\cdot\left(1+\dfrac{1}{x}\right)\left(1+\dfrac{1}{y}\right)\)
\(=\dfrac{\left(-y\right)\left(-x\right)}{xy}\cdot\left(1+\dfrac{1}{x}\right)\left(1+\dfrac{1}{y}\right)\)
\(=\left(1+\dfrac{1}{x}\right)\left(1+\dfrac{1}{y}\right)\)
\(=1+\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{xy}\ge1+\dfrac{4}{x+y}+\dfrac{1}{\dfrac{\left(x+y\right)^2}{4}}=1+\dfrac{4}{1}+\dfrac{1}{\dfrac{1}{4}}=9\)
Vậy \(B_{min}=9\Leftrightarrow x=y=\dfrac{1}{2}\)
Áp dụng BĐT Cauchy, ta có:
\(\left(1+\dfrac{1}{x}\right)^2+\left(1+\dfrac{1}{y}\right)^2\ge2\left(x+\dfrac{1}{x}\right)\left(x+\dfrac{1}{y}\right)\)
\(\Leftrightarrow............\ge2\left(xy+\dfrac{x}{y}+\dfrac{y}{x}+\dfrac{1}{xy}\right)\ge2\left(2\sqrt{xy\cdot\dfrac{1}{xy}}+2\sqrt{\dfrac{x}{y}\cdot\dfrac{y}{x}}\right)=2\cdot4=8\)
Vậy:.......
Cái này bổ sung, mk quên giải chung với cái kia
GTNN của A khi \(\left\{{}\begin{matrix}x+\dfrac{1}{x}=y+\dfrac{1}{y}\\x+y=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x^2+1}{x}=\dfrac{y^2+1}{y}\\x+y=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=y\\x+y=1\end{matrix}\right.\)\(\Leftrightarrow x=y=\dfrac{1}{2}\)
Vậy:GTNN của A là 8 khi x=y=1/2
Lời giải:
Áp dụng BĐT Cauchy ta có:
\(\frac{x^2}{y-1}+4(y-1)\geq 2\sqrt{4x^2}=4x\)
\(\frac{y^2}{x-1}+4(x-1)\geq 2\sqrt{4y^2}=4y\)
Cộng theo vế:
\(\frac{x^2}{y-1}+\frac{y^2}{x-1}+4(y-1)+4(x-1)\geq 4x+4y\)
\(\Leftrightarrow \frac{x^2}{y-1}+\frac{y^2}{x-1}\geq 8\)
Vậy \(P_{\min}=8\). Dấu bằng xảy ra khi \(x=y=2\)
Chắc đề đúng là số dương, vì ko tồn tại x;y nguyên dương thỏa mãn x+y=1
\(A=\dfrac{y^2}{xy+y}+\dfrac{x^2}{xy+x}\ge\dfrac{\left(x+y\right)^2}{x+y+2xy}\ge\dfrac{\left(x+y\right)^2}{x+y+\dfrac{1}{2}\left(x+y\right)^2}=\dfrac{2}{3}\)
Dấu "=" xảy ra khi \(x=y=\dfrac{1}{2}\)