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Áp dụng BĐT \(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\) ta có:
\(A=\frac{1}{1+a^2+b^2}+\frac{1}{2ab}\ge\frac{4}{1+a^2+b^2+2ab}\)
\(=\frac{4}{1+\left(a+b\right)^2}=\frac{4}{1+1}=2\)
Dấu "=" xảy ra khi \(\begin{cases}a=b\\a+b=1\end{cases}\)\(\Rightarrow a=b=\frac{1}{2}\)
Vậy \(Min_A=2\) khi \(a=b=\frac{1}{2}\)
\(P=\dfrac{a^2}{b^2}+\dfrac{b^2}{a^2}-\dfrac{2a}{b}-\dfrac{2b}{a}-1\)
Xét \(\dfrac{a}{a^2+1}+\dfrac{3\left(a-2\right)}{25}-\dfrac{2}{5}=\dfrac{a}{a^2+1}+\dfrac{3a-16}{25}=\dfrac{\left(3a-4\right)\left(a-2\right)^2}{25\left(a^2+1\right)}\ge0\)
\(\Rightarrow\dfrac{a}{a^2+1}\ge\dfrac{2}{5}-\dfrac{3\left(a-2\right)}{25}\)
CMTT \(\Rightarrow\left\{{}\begin{matrix}\dfrac{b}{b^2+1}\ge\dfrac{2}{5}-\dfrac{3\left(b-2\right)}{25}\\\dfrac{c}{c^2+1}\ge\dfrac{2}{5}-\dfrac{3\left(c-2\right)}{25}\end{matrix}\right.\)
Cộng vế theo vế:
\(\Rightarrow VT\ge\dfrac{2}{5}+\dfrac{2}{5}+\dfrac{2}{5}-\dfrac{3\left(a-2\right)+3\left(b-2\right)+3\left(c-2\right)}{25}\ge\dfrac{6}{5}-\dfrac{3\left(a+b+c-6\right)}{25}=\dfrac{6}{5}\)
Dấu \("="\Leftrightarrow a=b=c=2\)
Đặt \(\left(a+1;b+1;c+1\right)=\left(x;y;z\right)\Rightarrow1\le x\le y\le z\le2\)
\(B=\left(x+y+z\right)\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)=\dfrac{x}{y}+\dfrac{y}{z}+\dfrac{y}{x}+\dfrac{z}{y}+\dfrac{z}{x}+\dfrac{x}{z}+3\) (1)
Do \(x\le y\le z\Rightarrow\left(z-y\right)\left(y-x\right)\ge0\)
\(\Leftrightarrow xy+yz\ge y^2+zx\)
\(\Leftrightarrow\dfrac{x}{z}+1\ge\dfrac{y}{z}+\dfrac{x}{y}\)
Tương tự: \(1+\dfrac{z}{x}\ge\dfrac{y}{x}+\dfrac{z}{y}\)
Cộng vế: \(2+\dfrac{x}{z}+\dfrac{z}{x}\ge\dfrac{x}{y}+\dfrac{y}{z}+\dfrac{z}{y}+\dfrac{y}{x}\) (2)
Từ (1); (2) \(\Rightarrow B\le2\left(\dfrac{x}{z}+\dfrac{z}{x}\right)+5\)
Đặt \(\dfrac{z}{x}=t\Rightarrow1\le t\le2\)
\(\Rightarrow B\le2\left(t+\dfrac{1}{t}\right)+5=\dfrac{2t^2+2}{t}+5=\dfrac{2t^2+2}{t}-5+10\)
\(\Rightarrow B\le\dfrac{2t^2-5t+2}{t}+10=\dfrac{\left(t-2\right)\left(2t-1\right)}{t}+10\le10\)
\(B_{max}=10\) khi \(t=2\) hay \(\left(a;b;c\right)=\left(0;0;1\right);\left(0;1;1\right)\)
Áp dụng BĐT Minicopski, ta có:
\(P=\sqrt{a^2+\dfrac{1}{a^2}}+\sqrt{b^2+\dfrac{1}{b^2}}\ge\sqrt{\left(a+b\right)^2+\left(\dfrac{1}{a}+\dfrac{1}{b}\right)^2}\\ \Rightarrow P\ge\sqrt{4^2+\left(\dfrac{4}{a+b}\right)^2}=\sqrt{16+\left(\dfrac{4}{4}\right)^2}=\sqrt{17}\)
Đẳng thức xảy ra \(\Leftrightarrow a=b=2\)
Ta có \(a^2+\dfrac{1}{b+c}=a^2+\dfrac{1}{6-a}\)
Mà \(a+b+c=6\Rightarrow0\le a,b,c\le2\)
\(\Rightarrow a^2+\dfrac{1}{6-a}\ge2^2+\dfrac{1}{6-2}=\dfrac{17}{4}\)
\(\Rightarrow P=\sum\sqrt{a^2+\dfrac{1}{b+c}}=\sum\sqrt{a^2+\dfrac{1}{6-a}}\ge\sqrt{\dfrac{17}{4}}+\sqrt{\dfrac{17}{4}}+\sqrt{\dfrac{17}{4}}=\dfrac{3\sqrt{17}}{2}\)
Dấu \("="\Leftrightarrow a=b=c=2\)
Áp dụng bđt Cauchy-Schwarz:
\(P=\dfrac{1}{a^2+b^2}+\dfrac{1}{2ab}\ge\dfrac{\left(1+1\right)^2}{a^2+b^2+2ab}=\dfrac{4}{\left(a+b\right)^2}\ge\dfrac{4}{1^2}=4\)\("="\Leftrightarrow a=b=\dfrac{1}{2}\)