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a) Xét : \(P^2=\frac{3\left(a-b\right)^2}{3\left(a+b\right)^2}=\frac{3\left(a^2+b^2\right)-6ab}{3\left(a^2+b^2\right)+6ab}=\frac{10ab-6ab}{10ab+6ab}=\frac{4ab}{16ab}=\frac{1}{4}\)
Vì a > b > 0 nên P > 0 . Vậy \(P=\frac{1}{2}\)
b) Tương tự.
a/ \(3a^2+3b^2=10ab\Leftrightarrow3\left(a^2+b^2\right)=10ab\Leftrightarrow a^2+b^2=\frac{10ab}{3}\)
\(\Leftrightarrow a^2+b^2-2ab=\frac{10ab}{3}-2ab\Leftrightarrow\left(a-b\right)^2=\frac{4ab}{3}\)
tương tự: \(a^2+b^2=\frac{10ab}{3}\Leftrightarrow a^2+b^2+2ab=\frac{10ab}{3}+2ab\Leftrightarrow\left(a+b\right)^2=\frac{16ab}{3}\)
\(\Rightarrow P^2=\left(\frac{a-b}{a+b}\right)^2=\frac{\frac{4ab}{3}}{\frac{16ab}{3}}=\frac{1}{4}\Rightarrow P=\frac{1}{2}\)
Ap dung bdt \(\frac{1}{x+y}\le\frac{1}{4}\left(\frac{1}{x}+\frac{1}{y}\right).\left(x,y>0\right)\) lien tiep la duoc
Chuc bn thanh cong
svác-xơ ngược dấu.
\(\frac{16}{2a+3b+3c}=\frac{16}{\left(a+b\right)+\left(c+b\right)+\left(b+c\right)+\left(a+c\right)}\le\frac{1}{a+b}+\frac{2}{c+b}+\frac{1}{c+a}\)
Tương tự
\(\frac{16}{2b+3c+3a}\le\frac{1}{a+b}+\frac{1}{b+c}+\frac{2}{c+a}\)
\(\frac{16}{2c+3a+3b}\le\frac{2}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\)
Cộng lại ta được:
\(16VT\le4\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)\)
\(\Rightarrow VT\le\frac{1}{4}\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)\left(đpcm\right)\)
Áp dụng Cauchy, ta có:
\(a^4+b^2\ge2\sqrt{a^4b^2}=2a^2b\)
\(\Rightarrow\frac{1}{a^4+b^2+2ab^2}\le\frac{1}{2a^2b+2ab^2}\)
Tượng tự:
\(\frac{1}{b^4+a^2+2a^2b}\le\frac{1}{2a^2b+2ab^2}\)
\(\Rightarrow A\le\frac{2}{2ab\left(a+b\right)}\)
Lại có: \(\frac{1}{a}+\frac{1}{b}=2\)\(\Leftrightarrow\frac{a+b}{ab}=2\Rightarrow a+b=2ab\)
\(\Rightarrow A\le\frac{2}{\left(a+b\right)^2}\)
Áp dụng Schwarzt: \(2=\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\ge a+b\ge2\Rightarrow\left(a+b\right)^2\ge4\)
\(\Rightarrow A\le\frac{2}{4}=\frac{1}{2}\)
Dấu = xảy ra khi a=b=1
Áp dụng bđt cosi ta có :
A < = 1/2a^2b+2/ab^2 + 1/2ab^2+2a^2b
= 1/2ab . (1/a+b + 1/a+b) = 1/2ab . 2/a+b = 1/(a+b).(ab)
< = 1/\(\sqrt{ab}.2.ab\) = 1/2\(\sqrt{ab}^3\)
Có : 2 = 1/a + 1/b >= 2\(\sqrt{\frac{1}{ab}}\)
=> \(\sqrt{\frac{1}{ab}}\)< = 1
=> 1/ab < = 1
=> ab > =1
=> A < = 1/2.1 = 1/2
Dấu "=" xảy ra <=> a=b=1
Vậy GTLN của A = 1/2 <=> a=b=1
Tk mk nha
bố 32 tuổi
con 6 tuổi
ủng hộ nha
Câu b). Theo đầu bài ta có:
\(2a^2+2b^2=5ab\)
\(\Rightarrow2a^2+2b^2=ab+4ab\)
\(\Rightarrow2a^2+2b^2-4ab=ab\)
\(\Rightarrow2\left(a^2+b^2-2ab\right)=ab\)
\(\Rightarrow\left(a-b\right)^2=\frac{ab}{2}\)
\(\Rightarrow a-b=\sqrt{\frac{ab}{2}}\)
Mà \(2a^2+2b^2=5ab\)
\(\Rightarrow2a^2+2b^2=9ab-4ab\)
\(\Rightarrow2a^2+2b^2+4ab=9ab\)
\(\Rightarrow2\left(a^2+b^2+2ab\right)=9ab\)
\(\Rightarrow\left(a+b\right)^2=\frac{9ab}{2}\)
\(\Rightarrow a+b=\sqrt{\frac{9ab}{2}}\)
Từ trên suy ra:
\(Q=\frac{a+b}{a-b}=\left(a+b\right):\left(a-b\right)\)
\(\Leftrightarrow Q=\sqrt{\frac{9ab}{2}}:\sqrt{\frac{ab}{2}}\)
\(\Leftrightarrow Q=\sqrt{\frac{9ab}{2}:\frac{ab}{2}}\)
\(\Leftrightarrow Q=\sqrt{\frac{9\cdot ab\cdot2}{ab\cdot2}}\)
\(\Leftrightarrow Q=\sqrt{9}=3\)