Bài 2:

\(ab^2\le\dfrac{a^2+b^4}{2}\);   \(bc^2\le\dfrac{b^2+c^4}{2}\);   \(c\le\dfrac{c^2+1}{2}\);   \(a^2\le\dfrac{a^4+1}{2}\)

Cộng theo vế các BĐT trên, ta có:

\(VT\le\dfrac{1}{2}\left(a^4+b^4+c^4\right)+\dfrac{1}{2}\left(a^2+b^2+c^2\right)+1\)

\(\le\dfrac{1}{2}\left(a^4+b^4+c^4\right)+\dfrac{1}{2}\sqrt{3\left(a^4+b^4+c^4\right)}+1\)

\(=\dfrac{3}{2}+\dfrac{3}{2}+1=4\)

Đẳng thức xảy ra\(\Leftrightarrow a=b=c=1\)

Câu 1:

Đặt \(n^2=k\left(k\ge0\right)\)

Ta có:

\(M=4^k+2^{k+2}-5=\left(2^k\right)^2+2.2^k.2+4-9\)

\(=\left(2^k+2\right)^2-3^2=\left(2^k-1\right)\left(2^k+5\right)\)

Để M là SNT thì \(2^k-1=1\) và \(2^k+5\) là SNT

\(\Rightarrow k=1\Leftrightarrow\left[{}\begin{matrix}n=1\\n=-1\end{matrix}\right.\)(tm)

Câu 2:

Ta có:

\(\left(x^2+xy+y^2\right)⋮10\Rightarrow\left[\left(x-y\right)\left(x^2+xy+y^2\right)\right]⋮10\)

\(\Leftrightarrow x^3\equiv y^3\left(mod10\right)\Leftrightarrow x\equiv y\left(mod10\right)\)

Theo gt thì

\(x^2+xy+y^2\equiv0\left(mod10\right)\Rightarrow3x^2\equiv0\left(mod10\right)\)

\(\Leftrightarrow x^2\equiv0\left(mod10\right)\)

\(\Leftrightarrow x\equiv0\left(mod10\right)\Rightarrow\left(x^2+xy+y^2\right)⋮100\)(đpcm)

Chứng minh BĐT phụ \(a^3+b^3\ge ab\left(a+b\right)\)

Áp dụng BĐT phụ ta có:

\(a^3+b^3+c^3\ge\dfrac{1}{2}\left(a^2b+ab^2+a^2c+ac^2+b^2c+bc^2\right)\)

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

\(\ge a^2\sqrt{bc}+b^2\sqrt{ac}+c^2\sqrt{ab}\)

Đẳng thức xảy ra khi và chỉ khi \(a=b=c\)

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Ta có:

\(A^2=x-2+\sqrt{y-3}=6-y-2+\sqrt{y-3}\)

\(=4-y+\sqrt{y-3}\)

Đặt \(\sqrt{y-3}=t\) \(\left(t\ge0\right)\)

\(\Rightarrow A^2=4-t^2-3+t=-t^2+t+1\)

Khi đó \(Max_{A^2}=\dfrac{5}{4}\Leftrightarrow t=\dfrac{1}{2}\)

\(\Rightarrow Max_A=\dfrac{\sqrt{5}}{2}\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{11}{4}\\y=\dfrac{13}{4}\end{matrix}\right.\)(tm)

GT \(\Rightarrow\left(x_1-\dfrac{3}{2}\right)\left(x_2-\dfrac{3}{2}\right)< 0\)

\(\Leftrightarrow x_1x_2-\dfrac{3}{2}\left(x_1+x_2\right)+\dfrac{9}{4}< 0\)

Đến đây dùng hệ thức Vi ét là ra

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