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Ta có:\(A=\dfrac{1}{ab}+\dfrac{1}{a^2+b^2}\)
\(A=\dfrac{1}{2ab}+\dfrac{1}{2ab}+\dfrac{1}{a^2+b^2}\)
\(A\ge\dfrac{1}{\dfrac{\left(a+b\right)^2}{2}}+\dfrac{4}{a^2+2ab+b^2}\)
\(A\ge2+4=6\)
"="<=>a=b=0,5
Vậy MINA=6<=>a=b=0,5
Theo C.B.S thì
\(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ac}\ge\dfrac{9}{ab+bc+ac}\)
\(\Rightarrow\dfrac{1}{a^2+b^2+c^2}+\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ac}\ge\dfrac{1}{a^2+b^2+c^2}+\dfrac{9}{ab+bc+ac}=\dfrac{1}{a^2+b^2+c^2}+\dfrac{1}{ab+bc+ac}+\dfrac{1}{ab+bc+ac}+\dfrac{7}{ab+bc+ac}\)
Lại theo CBS thì
\(\dfrac{1}{a^2+b^2+c^2}+\dfrac{1}{ab+bc+ac}+\dfrac{1}{ab+bc+ac}\ge\dfrac{9}{\left(a+b+c\right)^2}=9\)mà \(ab+bc+ac\le\dfrac{\left(a+b+c\right)^2}{3}=\dfrac{1}{3}\)
\(\Rightarrow\dfrac{7}{ab+bc+ac}\ge21\)
\(\Rightarrow\)\(\dfrac{1}{a^2+b^2+c^2}+\dfrac{1}{ab+bc+ac}+\dfrac{1}{ab+bc+ac}+\dfrac{7}{ab+bc+ac}\)\(\)\(\ge21+9=30\)
vậy Min = 30 khi a = b = c = 1/3
Bài 1:
Áp dụng BĐT Cauchy-Schwarz:
\(\frac{1}{2ab}+\frac{1}{a^2+b^2}\geq \frac{4}{2ab+a^2+b^2}=\frac{4}{a+b)^2}=4(1)\)
Áp dụng BĐT AM-GM:
\(1=a+b\geq 2\sqrt{ab}\Rightarrow ab\leq \frac{1}{4}\Rightarrow \frac{3}{2ab}\geq 6(2)\)
\(a^4+b^4\geq \frac{(a^2+b^2)^2}{2}\geq \frac{(\frac{(a+b)^2}{2})^2}{2}=\frac{1}{8}\) \(\Rightarrow \frac{a^4+b^4}{2}\geq \frac{1}{16}(3)\)
Từ \((1);(2);(3)\Rightarrow P\geq 4+6+\frac{1}{16}=\frac{161}{16}\)
Vậy \(P_{\min}=\frac{161}{16}\). Dấu bằng xảy ra tại $a=b=0,5$
Bài 2:
Áp dụng BĐT Cauchy-Schwarz:
\(2\left(\frac{1}{x^2+y^2}+\frac{1}{2xy}\right)\geq 2. \frac{4}{x^2+y^2+2xy}=\frac{8}{(x+y)^2}=\frac{9}{2}\)
Áp dụng BĐT AM-GM:
\(\frac{80}{81xy}+5xy\geq 2\sqrt{\frac{80}{81}.5}=\frac{40}{9}\)
\(\frac{4}{3}=a+b\geq 2\sqrt{ab}\Rightarrow ab\leq \frac{4}{9}\Rightarrow \frac{1}{81ab}\geq \frac{1}{36}\)
Cộng những BĐT vừa cm được ở trên với nhau:
\(\Rightarrow A\geq \frac{9}{2}+\frac{40}{9}+\frac{1}{36}=\frac{323}{36}\)
Vậy \(A_{\min}=\frac{323}{36}\Leftrightarrow a=b=\frac{2}{3}\)
Áp dụng BĐT Cauchy cho các số dương , ta có :
\(a+\dfrac{1}{4a}\text{ ≥}2\sqrt{a.\dfrac{1}{4a}}=2.\dfrac{1}{2}=1\)
\(b+\dfrac{1}{4b}\text{ ≥}2\sqrt{b.\dfrac{1}{4b}}=2.\dfrac{1}{2}=1\)
\(c+\dfrac{1}{4c}\text{ ≥}2\sqrt{c.\dfrac{1}{4c}}=2.\dfrac{1}{2}=1\)
⇒ \(a+b+c+\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\text{ ≥}3\)
⇔ \(a+b+c+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\text{ ≥}3+\dfrac{3}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\text{ ≥ }3+\dfrac{3}{4}.\dfrac{\left(1+1+1\right)^2}{a+b+c}=3+\dfrac{3}{4}.\dfrac{9}{a+b+c}\text{ ≥}3+\dfrac{3}{4}.\dfrac{9}{\dfrac{3}{2}}=\dfrac{15}{2}\) ⇒ \(A_{MIN}=\dfrac{15}{2}."="\text{⇔}a=b=c=\dfrac{1}{2}\)
☘ Áp dụng bất đửng thức AM - GM
\(\Rightarrow A=\left(a+b+1\right)\left(a^2+b^2\right)+\dfrac{4}{a+b}\)
\(\ge\left(a+b+1\right)\times2ab+\dfrac{4}{a+b}\)
\(=2\left(a+b+1\right)+\dfrac{4}{a+b}\)
\(=\left(a+b+\dfrac{4}{a+b}\right)+\left(a+b\right)+2\)
\(\ge4+2\sqrt{ab}+2=8\)
⚠ Tự kết luận nha.
\(H=\sqrt{a^2+\dfrac{1}{b^2}}+\sqrt{b^2+\dfrac{1}{c^2}}+\sqrt{c^2+\dfrac{1}{a^2}}\)
\(\ge\sqrt{\left(a+b+c\right)^2+\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2}\)
\(\ge\sqrt{\left(a+b+c\right)^2+\dfrac{81}{\left(a+b+c\right)^2}}\)
\(\ge\sqrt{\left(\dfrac{3}{2}\right)^2+\dfrac{81}{\left(\dfrac{3}{2}\right)^2}}=\dfrac{3\sqrt{17}}{2}\)
Lời giải:
Áp dụng BĐT Cauchy-Schwarz:
$P=\frac{18}{a^2+b^2}+\frac{10}{2ab}\geq \frac{(\sqrt{18}+\sqrt{10})^2}{a^2+b^2+2ab}$
$=\frac{(\sqrt{18}+\sqrt{10})^2}{(a+b)^2}=(\sqrt{18}+\sqrt{10})^2=28+12\sqrt{5}$
Vậy $P_{\min}=28+12\sqrt{5}$