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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
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 \(x=y=z=1\)
Ta sẽ cm bđt:\(\dfrac{x^2}{a}+\dfrac{y^2}{b}+\dfrac{z^2}{c}\ge\dfrac{\left(x+y+z\right)^2}{a+b+c}\)
Áp dụng bđt bunhia:
\(\left(\dfrac{x^2}{a}+\dfrac{y^2}{b}+\dfrac{z^2}{c}\right)\left(a+b+c\right)\ge\left(x+y+z\right)^2\)
\(\Rightarrow\dfrac{x^2}{a}+\dfrac{y^2}{b}+\dfrac{z^2}{c}\ge\dfrac{\left(x+y+z\right)^2}{a+b+c}\)
Áp dụng vào suy ra:
\(A=\dfrac{x^2}{y+z}+\dfrac{y^2}{z+x}+\dfrac{z^2}{x+y}\ge\dfrac{\left(x+y+z\right)^2}{2\left(x+y+z\right)}=\dfrac{2}{2}=1\)
"="<=>x=y=z=\(\dfrac{2}{3}\)
Cách khác:
Áp dụng BĐT Cauchy cho các số dương ta có:
\(\frac{x^2}{y+z}+\frac{y+z}{2}\geq 2\sqrt{\frac{x^2}{y+z}.\frac{y+z}{4}}=x\)
\(\frac{y^2}{x+z}+\frac{x+z}{4}\geq 2\sqrt{\frac{y^2}{x+z}.\frac{x+z}{4}}=y\)
\(\frac{z^2}{x+y}+\frac{x+y}{4}\geq 2\sqrt{\frac{z^2}{x+y}.\frac{x+y}{4}}=z\)
Cộng theo vế và rút gọn ta có:
\(A+\frac{x+y+z}{2}\geq x+y+z\)
\(\Rightarrow A\geq \frac{x+y+z}{2}=1\)
Vậy \(A_{\min}=1\Leftrightarrow x=y=z=\frac{2}{3}\)
Lời giải:
Áp dụng BĐT Bunhiacopxky:
\(\left(\frac{2}{x}+\frac{8}{9y}+\frac{18}{25z}\right)(x+y+z)\geq (\sqrt{2}+\sqrt{\frac{8}{9}}+\sqrt{\frac{18}{25}})^2\)
$\Leftrightarrow A.2\geq \frac{2312}{225}$
$\Leftrightarrow A\geq \frac{1156}{225}$
Vậy $A_{\min}=\frac{1156}{225}$