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\(ab\le\frac{a^2+b^2}{2}\le\frac{16}{2}=8\)
Ta có: \(N^2=\left(a\sqrt{9b\left(a+8b\right)}+b\sqrt{9a\left(b+8a\right)}\right)^2\)
\(\le\left(a^2+b^2\right)\left[9b\left(a+8b\right)+9a\left(b+8a\right)\right]\)
\(\le16\left(18ab+72\left(a^2+b^2\right)\right)\le16\left(18.8+72.16\right)\)
\(=20736\)
=> \(N\le144\)
Dấu "=" xảy ra <=> a = b = \(\sqrt{8}\)
Vậy max N = 144 tại a = b = \(\sqrt{8}\)
Ta có \(2=a^2+b^2\ge2ab\)
\(\Leftrightarrow ab\le1\)
\(M\le\sqrt{\left(a^2+b^2\right)\left(36ab+45b^2+36ab+45a^2\right)}\)
\(=\sqrt{2\left(72ab+90\right)}\)\(\le\sqrt{2\left(72+90\right)}=\sqrt{324}=18\)
GTLN là 18 đạt được khi a = b = 1
Ta có \(2ab\le a^2+b^2\)
Áp dụng BĐT Bunhia:
\(M^2\le\left(a^2+b^2\right)\left(b\left(a+8b\right)+a\left(b+8a\right)\right)\)
\(\Rightarrow M^2\le\left(a^2+b^2\right)\left(2ab+8b^2+8a^2\right)\)
\(\Rightarrow M^2\le\left(a^2+b^2\right)\left(8a^2+8b^2+a^2+b^2\right)\)
\(\Rightarrow M^2\le9\left(a^2+b^2\right)^2\Rightarrow M\le3\left(a^2+b^2\right)=48\)
\(\Rightarrow M_{max}=48\) khi \(a=b=2\sqrt{2}\)
\(P=\sqrt{a\left(b+1\right)}+\sqrt{b\left(a+1\right)}\)
\(\Rightarrow P\sqrt{2}=\sqrt{2a\left(b+1\right)}+\sqrt{2b\left(a+1\right)}\)
\(\le\frac{1}{2}\left(2a+b+1\right)+\frac{1}{2}\left(2b+a+1\right)\)
\(\le\frac{1}{2}\left(3a+3b+2\right)\le\frac{1}{2}.\left(3.2+2\right)=4\)
\(\Rightarrow p\le2\sqrt{2}\)
Dấu"=" xảy ra \(\Leftrightarrow a=b=1\)
Vậy Max P \(=2\sqrt{2}\)\(\Leftrightarrow a=b=1\)
Bài 1:
Áp dụng BĐT Bunhiacopxky:
\(M^2=(a\sqrt{9b(a+8b)}+b\sqrt{9a(b+8a)})^2\)
\(\leq (a^2+b^2)(9ab+72b^2+9ab+72a^2)\)
\(\Leftrightarrow M^2\leq (a^2+b^2)(72a^2+72b^2+18ab)\)
Áp dụng BĐT AM-GM: \(a^2+b^2\geq 2ab\Rightarrow 18ab\leq 9(a^2+b^2)\)
Do đó, \(M^2\leq (a^2+b^2)(72a^2+72b^2+9a^2+9b^2)=81(a^2+b^2)^2\)
\(\Leftrightarrow M\leq 9(a^2+b^2)\leq 144\)
Vậy \(M_{\max}=144\Leftrightarrow a=b=\sqrt{8}\)
Bài 6:
\(a+\frac{1}{a-1}=1+(a-1)+\frac{1}{a-1}\)
Vì \(a>1\rightarrow a-1>0\). Do đó áp dụng BĐT Am-Gm cho số dương\(a-1,\frac{1}{a-1}\) ta có:
\((a-1)+\frac{1}{a-1}\geq 2\sqrt{\frac{a-1}{a-1}}=2\)
\(\Rightarrow a+\frac{1}{a-1}=1+(a-1)+\frac{1}{a-1}\geq 3\) (đpcm)
Dấu bằng xảy ra khi \(a-1=1\Leftrightarrow a=2\)
Bài 3:
Xét \(\sqrt{a^2+1}\). Vì \(ab+bc+ac=1\) nên:
\(a^2+1=a^2+ab+bc+ac=(a+b)(a+c)\)
\(\Rightarrow \sqrt{a^2+1}=\sqrt{(a+b)(a+c)}\)
Áp dụng BĐT AM-GM có: \(\sqrt{(a+b)(a+c)}\leq \frac{a+b+a+c}{2}=\frac{2a+b+c}{2}\)
hay \(\sqrt{a^2+1}\leq \frac{2a+b+c}{2}\)
Hoàn toàn tương tự với các biểu thức còn lại và cộng theo vế:
\(\sqrt{a^2+1}+\sqrt{b^2+1}+\sqrt{c^2+1}\leq \frac{2a+b+c}{2}+\frac{2b+a+c}{2}+\frac{2c+a+b}{2}=2(a+b+c)\)
Ta có đpcm. Dấu bằng xảy ra khi \(a=b=c=\frac{1}{\sqrt{3}}\)
Bài 4:
Ta có:
\(A=\frac{8a^2+b}{4a}+b^2=2a+\frac{b}{4a}+b^2\)
\(\Leftrightarrow A+\frac{1}{4}=2a+\frac{b+a}{4a}+b^2=2a+b+\frac{b+a}{4a}+b^2-b\)
Vì \(a+b\geq 1, a>0\) nên \(A+\frac{1}{4}\geq a+1+\frac{1}{4a}+b^2-b\)
Áp dụng BĐT AM-GM:
\(a+\frac{1}{4a}\geq 2\sqrt{\frac{1}{4}}=1\)
\(\Rightarrow A+\frac{1}{4}\geq 2+b^2-b=\left(b-\frac{1}{2}\right)^2+\frac{7}{4}\geq \frac{7}{4}\)
\(\Leftrightarrow A\geq \frac{3}{2}\).
Vậy \(A_{\min}=\frac{3}{2}\Leftrightarrow a=b=\frac{1}{2}\)
a) \(A=\left(\sqrt{a}+\sqrt{b}\right)^2\le\left(\sqrt{a}+\sqrt{b}\right)^2+\left(\sqrt{a}-\sqrt{b}\right)^2=2a+2b\le2\)
Vậy GTLN của A là 2 \(\Leftrightarrow\hept{\begin{cases}\sqrt{a}=\sqrt{b}\\a+b=1\end{cases}\Leftrightarrow a=b=\frac{1}{2}}\)
b) Ta có : \(\left(\sqrt{a}+\sqrt{b}\right)^4\le\left(\sqrt{a}+\sqrt{b}\right)^4+\left(\sqrt{a}-\sqrt{b}\right)^4=2\left(a^2+b^2+6ab\right)\)
Tương tự : \(\left(\sqrt{a}+\sqrt{c}\right)^4\le2\left(a^2+c^2+6ac\right)\)
\(\left(\sqrt{a}+\sqrt{d}\right)^4\le2\left(a^2+d^2+6ad\right)\)
\(\left(\sqrt{b}+\sqrt{c}\right)^4\le2\left(b^2+c^2+6bc\right)\)
\(\left(\sqrt{b}+\sqrt{d}\right)^4\le2\left(b^2+d^2+6bd\right)\)
\(\left(\sqrt{c}+\sqrt{d}\right)^4\le2\left(c^2+d^2+6cd\right)\)
Cộng các vế lại, ta được :
\(B\le6\left(a^2+b^2+c^2+d^2+2ab+2ac+2ad+2bd+2cd+2bc\right)=6\left(a+b+c+d\right)^2\)
\(\Rightarrow B\le6\)
Vậy GTLN của B là 6 \(\Leftrightarrow\hept{\begin{cases}\sqrt{a}=\sqrt{b}=\sqrt{c}=\sqrt{d}\\a+b+c+d=1\end{cases}}\Leftrightarrow a=b=c=d=\frac{1}{4}\)
By C-S and AM-GM's inequality
\(M=a\left(9b\left(a+8b\right)\right)^{\dfrac{1}{2}}+b\left(9a\left(b+8a\right)\right)^{\dfrac{1}{2}}\)
\(\le\left(\left(a^2+b^2\right)\left(9b\left(a+8b\right)+9a\left(b+8a\right)\right)\right)^{\dfrac{1}{2}}\)
\(=\left(\left(a^2+b^2\right)\left(18ab+72b^2+72a^2\right)\right)^{\dfrac{1}{2}}\)
\(=\left(\left(a^2+b^2\right)\left(18\cdot\dfrac{a^2+b^2}{2}+72b^2+72a^2\right)\right)^{\dfrac{1}{2}}\)
\(=\left(16\cdot\left(18\cdot\dfrac{16}{2}+72\cdot16\right)\right)^{\dfrac{1}{2}}=144\)
\("="\Leftrightarrow a=b=2\sqrt{2}\)