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Ảo loz ak

Ta có \(a^2+b^2\ge2ab\Rightarrow2ab\le1\Rightarrow ab\le\dfrac{1}{2}\)( áp dụng cosi cho 2 số dương)

Ta có BĐT \(\dfrac{\left(a+b\right)^2}{4}\le\dfrac{a^2+b^2}{2}\)(*)

Thật vậy (*)\(\left(a-b\right)^2\ge0\)(luôn đúng)

\(\Rightarrow\left(a+b\right)^2\le2\left(a^2+b^2\right)\)\(\Leftrightarrow2\left(a+b\right)\le2\sqrt{2\left(a^2+b^2\right)}=2\sqrt{2}\)

Vậy S=ab+2(a+b)\(\le2\sqrt{2}+\dfrac{1}{2}=\dfrac{4\sqrt{2}+1}{2}\)

Dấu "=" xảy ra \(\Leftrightarrow a=b=\dfrac{\sqrt{2}}{2}\)

\(S=ab+2\left(a+b\right)\le\dfrac{1}{2}\left(a^2+b^2\right)+2\sqrt{2\left(a^2+b^2\right)}=\dfrac{1}{2}+2\sqrt{2}\)

Dấu "=" xảy ra khi \(a=b=\dfrac{\sqrt{2}}{2}\)

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

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

\(S=\frac{\left\{\left(a+b\right)-2\right\}^2+5}{2}\)

S>=\(\frac{5}{2}\) xay ra dau = khi va chi khi a+b=2 dua vao day tim a,b

\(5,M=a^3+b^3=\left(a+b\right)\left(a^2-ab+b^2\right)\\ M=\left(a+b\right)\left[\left(a+b\right)^2-3ab\right]\\ M=1\left(1-3ab\right)=1-3ab\ge1-\dfrac{3\left(a+b\right)^2}{4}=1-\dfrac{3}{4}=\dfrac{1}{4}\\ M_{min}=\dfrac{1}{4}\Leftrightarrow a=b=\dfrac{1}{2}\)

Câu 5:

\(a+b=1\Rightarrow a=1-b\)

\(M=a^3+b^3=\left(1-b\right)^3+b^3=1-3b+3b^2-b^3+b^3\)

\(=1-3b+3b^2=3\left(b^2-b+\dfrac{1}{4}\right)+\dfrac{1}{4}=3\left(b-\dfrac{1}{2}\right)^2+\dfrac{1}{4}\ge\dfrac{1}{4}\)

\(minM=\dfrac{1}{4}\Leftrightarrow a=b=\dfrac{1}{2}\)

Câu 7:

\(a^3+b^3+abc\ge ab\left(a+b+c\right)\)

\(\Leftrightarrow a^3+b^3+abc-ab\left(a+b+c\right)\ge0\)

\(\Leftrightarrow a^3+b^3-a^2b-ab^2\ge0\)

\(\Leftrightarrow a^2\left(a-b\right)-b^2\left(a-b\right)\ge0\)

\(\Leftrightarrow\left(a-b\right)\left(a^2-b^2\right)\ge0\Leftrightarrow\left(a-b\right)^2\left(a+b\right)\ge0\)(đúng do a,b dương)

Dấu "=" xảy ra \(\Leftrightarrow a=b\)

giúp mình vs

5.

Với mọi a;b ta có: \(\left(a-b\right)^2\ge0\Rightarrow a^2+b^2\ge2ab\Rightarrow2a^2+2b^2\ge a^2+b^2+2ab\)

\(\Rightarrow2\left(a^2+b^2\right)\ge\left(a+b\right)^2\Rightarrow a^2+b^2\ge\dfrac{1}{2}\left(a+b\right)^2=\dfrac{1}{2}\)

\(M=a^3+b^3=\left(a+b\right)\left(a^2+b^2-ab\right)=a^2+b^2-ab\)

\(M=\dfrac{3}{2}\left(a^2+b^2\right)-\dfrac{1}{2}\left(a+b\right)^2=\dfrac{3}{2}\left(a^2+b^2\right)-\dfrac{1}{2}\ge\dfrac{3}{2}.\dfrac{1}{2}-\dfrac{1}{2}=\dfrac{1}{4}\)

\(M_{min}=\dfrac{1}{4}\) khi \(a=b=\dfrac{1}{2}\)

6.

Do \(a^3+b^3=\left(a+b\right)\left(a^2-ab+b^2\right)=2>0\)

Mà \(a^2-ab+b^2>0\Rightarrow a+b>0\)

Mặt khác với mọi a;b ta có:

\(\left(a-b\right)^2\ge0\Rightarrow a^2+b^2\ge2ab\Rightarrow a^2+b^2+2ab\ge4ab\)

\(\Rightarrow\left(a+b\right)^2\ge4ab\Rightarrow ab\le\dfrac{1}{4}\left(a+b\right)^2\) \(\Rightarrow-ab\ge-\dfrac{1}{4}\left(a+b\right)^2\)

Từ đó:

\(2=a^3+b^3=\left(a+b\right)^3-3ab\left(a+b\right)\ge\left(a+b\right)^3-3.\dfrac{1}{4}\left(a+b\right)^2\left(a+b\right)=\dfrac{1}{4}\left(a+b\right)^3\)

\(\Rightarrow\left(a+b\right)^3\le8\Rightarrow a+b\le2\)

\(N_{max}=2\) khi \(a=b=1\)

7.

Ta có:

\(a^3+b^3+abc=\left(a+b\right)\left(a^2+b^2-ab\right)+abc\ge\left(a+b\right)\left(2ab-ab\right)+abc\)

\(\Rightarrow a^3+b^3+abc\ge ab\left(a+b\right)+abc=ab\left(a+b+c\right)\) (đpcm)

Dấu "=" xảy ra khi \(a=b\)

8.

\(\left|a+b\right|>\left|a-b\right|\Leftrightarrow\left(a+b\right)^2>\left(a-b\right)^2\)

\(\Leftrightarrow a^2+2ab+b^2>a^2-2ab+b^2\)

\(\Leftrightarrow4ab>0\Leftrightarrow ab>0\)

\(\Rightarrow a;b\) cùng dấu

so am dc k.o ban

a,b thực phải không

Đặt \(a^2+b^2=x\)

Ta có : a + b = 2 \(\Rightarrow a^2+b^2+2ab=4\Rightarrow ab=2-\frac{x}{2}\)

\(\Rightarrow A=x\left(2-\frac{x}{2}\right)=2x-\frac{x^2}{2}=2-\frac{1}{2}\left(x-2\right)^2\le2\)

Vậy GTLN của P là 2 \(\Leftrightarrow\)a =b = 1

\(A=ab\left(a^2+b^2\right)=\frac{1}{8}\left(a+b\right)^4-\frac{1}{8}\left(a-b\right)^4\le\frac{1}{8}\left(a+b\right)^4=2\)

Đẳng thức xảy ra khi a = b = 1