\(2a^2+\frac{1}{a^2}+\frac{b^2}{4}=4\Leftrightarrow\left(a^2+\frac{1}{a^2}-2\right)+\left(a^2+\frac{b^2}{4}-ab\right)=4-ab-2\)

\(\Leftrightarrow\left(a-\frac{1}{a}\right)^2+\left(a-\frac{b}{2}\right)^2=2-ab\)

\(VF=2-ab=\left(a-\frac{1}{a}\right)^2+\left(b-\frac{b}{2}\right)^2\ge0\)

Hay \(ab\le2\)

Dấu "=" xảy ra khi \(\hept{\begin{cases}a=\frac{1}{a}\\b=\frac{b}{2}\end{cases}}\Leftrightarrow\orbr{\begin{cases}\left(a;b\right)=\left(1;\frac{1}{2}\right)\\\left(a;b\right)=\left(-1;-\frac{1}{2}\right)\end{cases}}\)

ủa bạn tìm giá trị nhỏ nhất của biểu thức S=ab+2019 mà 

a) 

b) 

https://hoc24.vn/cau-hoi/c-voi-a-b-c-la-cac-so-duong-thoa-man-dieu-kien-a-b-c-2-tim-max-q-sqrt2abcsqrt2bcasqrt2cab.8298826302

Bạn có thể tham khảo ở đây. Đừng quên like giúp mik nha bạn. Thx

Ta có \(\frac{x^2+y^2+z^2}{a^2+b^2+c^2}=\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}\)

\(\Leftrightarrow\frac{x^2}{a^2+b^2+c^2}+\frac{y^2}{a^2+b^2+c^2}+\frac{z^2}{a^2+b^2+c^2}=\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}\)

\(\Leftrightarrow\frac{x^2}{a^2+b^2+c^2}-\frac{x^2}{a^2}+\frac{y^2}{a^2+b^2+c^2}-\frac{y^2}{b^2}+\frac{z^2}{a^2+b^2+c^2}-\frac{z^2}{c^2}=0\)

\(\Leftrightarrow x^2\left(\frac{1}{a^2+b^2+c^2}-\frac{1}{a^2}\right)+y^2\left(\frac{1}{a^2+b^2+c^2}-\frac{1}{b^2}\right)+z^2\left(\frac{1}{a^2+b^2+c^2}-\frac{1}{c^2}\right)=0\)

Do \(\left\{\begin{matrix}\frac{1}{a^2+b^2+c^2}-\frac{1}{a^2}\\\frac{1}{a^2+b^2+c^2}-\frac{1}{b^2}\\\frac{1}{a^2+b^2+c^2}-\frac{1}{c^2}\end{matrix}\right.\ne0\) và \(a,b,c\ne0\)

\(\Rightarrow\left\{\begin{matrix}x^2=0\\y^2=0\\z^2=0\end{matrix}\right.\)

\(\Rightarrow\left\{\begin{matrix}x=0\\y=0\\z=0\end{matrix}\right.\)

Ta có \(A=x^{2008}+y^{2008}+z^{2008}\)

\(\Rightarrow A=0+0+0\)

\(\Rightarrow A=0\)

Vậy A = 0

Áp dụng bổ đề quen thuộc \(x^3+y^3\ge xy\left(x+y\right)\), ta được: \(\frac{1}{2a^3+b^3+c^3+2}=\frac{1}{\left(a^3+b^3\right)+\left(a^3+c^3\right)+2}\le\frac{1}{ab\left(a+b\right)+ac\left(a+c\right)+2}\)\(=\frac{bc}{ab^2c\left(a+b\right)+abc^2\left(a+c\right)+2bc}=\frac{bc}{b\left(a+b\right)+c\left(a+c\right)+2bc}\)\(\le\frac{bc}{ab+ac+4bc}=\frac{bc}{b\left(a+c\right)+c\left(a+b\right)+2bc}\)\(\le\frac{1}{9}\left(\frac{bc}{b\left(a+c\right)}+\frac{bc}{c\left(a+b\right)}+\frac{bc}{2bc}\right)=\frac{1}{9}\left(\frac{c}{a+c}+\frac{b}{a+b}+\frac{1}{2}\right)\)(1)

Tương tự, ta có: \(\frac{1}{a^3+2b^3+c^3+2}\le\frac{1}{9}\left(\frac{c}{b+c}+\frac{a}{a+b}+\frac{1}{2}\right)\)(2); \(\frac{1}{a^3+b^3+2c^3+2}\le\frac{1}{9}\left(\frac{b}{b+c}+\frac{a}{a+c}+\frac{1}{2}\right)\)(3)

Cộng theo vế ba bất đẳng thức (1), (2), (3), ta được: \(P\le\frac{1}{9}\left(1+1+1+\frac{3}{2}\right)=\frac{1}{2}\)

Vậy giá trị lớn nhất của P là \(\frac{1}{2}\)đạt được khi x = y = z = 1

Từ \(2a+2b+2c=3abc\)

\(\Leftrightarrow\frac{2}{3bc}+\frac{2}{3ac}+\frac{2}{3ab}=1\left(1\right)\)

Khi đó \(P=\frac{b}{a^2}+\frac{c}{b^2}+\frac{a}{c^2}-\frac{2}{a^2}-\frac{2}{b^2}-\frac{2}{c^2}\)

Áp dụng BĐT AM-GM ta có: 

\(\frac{b}{a^2}+\frac{c}{b^2}+\frac{a}{c^2}\ge3\sqrt[3]{\frac{b}{a^2}\cdot\frac{c}{b^2}\cdot\frac{a}{c^2}}=3\sqrt[3]{\frac{1}{abc}}\)

\(P_{Min}\) xảy ra khi \(\frac{b}{a^2}+\frac{c}{b^2}+\frac{a}{c^2}=3\sqrt[3]{\frac{1}{abc}}\forall a=b=c\left(2\right)\)

Từ \(\left(1\right);\left(2\right)\Rightarrow a=b=c=\sqrt{2}\)

Khi đó \(P_{Min}=3\sqrt[3]{\frac{1}{abc}}-\frac{2}{a^2}-\frac{2}{b^2}-\frac{2}{c^2}=\frac{3\sqrt{2}-6}{2}\)

Đẳng thức xảy ra khi \(a=b=c=\sqrt{2}\)

Bài này giải như này cơ:

\(2a+2b+2c=3abc\)\(\Rightarrow\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=\frac{3}{2}\)

\(P=\frac{\left(a-1\right)+\left(b-1\right)}{a^2}+\frac{\left(b-1\right)+\left(c-1\right)}{b^2}+\frac{\left(c-1\right)+\left(a-1\right)}{c^2}-\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)

\(=\left(a-1\right)\left(\frac{1}{a^2}+\frac{1}{c^2}\right)+\left(b-1\right)\left(\frac{1}{a^2}+\frac{1}{b^2}\right)+\left(c-1\right)\left(\frac{1}{b^2}+\frac{1}{c^2}\right)-\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)

\(\ge\frac{2\left(a-1\right)}{ac}+\frac{2\left(b-1\right)}{ab}+\frac{2\left(c-1\right)}{bc}-\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)

\(=\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)=\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-3\)

\(\ge\sqrt{3\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)}-3=\sqrt{3.\frac{3}{2}}-3=\frac{3\sqrt{2}-6}{2}\)

Vậy \(minP=\frac{3\sqrt{2}-6}{2}\Leftrightarrow a=b=c=\sqrt{2}\)

cả 2 cách đều hay :*

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dễ thì bn giải hộ mk đi,nói đc lm đc nhỉ