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\(A=\frac{4ab}{a^2-b^2}=\frac{4.\frac{a}{b}}{\left(\frac{a}{b}\right)^2-1}\Leftrightarrow A\left(\frac{a}{b}\right)^2-4\frac{a}{b}-A=0\Leftrightarrow At^2-4t+\frac{4}{A}=A+\frac{4}{A}\)
\(t=\frac{2}{A^2}+-\sqrt{\frac{A^2+4}{A^3}}\)
\(B=\frac{4a^4b^4}{a^8-b^8}=\frac{4t^4}{t^8-1}=..\)
a/
\(VT\ge\frac{\frac{1}{2}\left(a+b\right)^2}{a+b}+\frac{\frac{1}{2}\left(b+c\right)^2}{b+c}+\frac{\frac{1}{2}\left(c+a\right)^2}{c+a}=a+b+c\ge3\sqrt[3]{abc}=3\)
Dấu "=" xảy ra khi \(a=b=c=1\)
b/ Ta có: \(x^4+y^4\ge\frac{1}{2}\left(x^2+y^2\right)\left(y^2+y^2\right)\ge xy\left(x^2+y^2\right)\)
\(\Rightarrow VT\le\frac{1}{a+bc\left(b^2+c^2\right)}+\frac{1}{b+ca\left(a^2+c^2\right)}+\frac{1}{c+ab\left(a^2+b^2\right)}\)
\(VT\le\frac{1}{a+\frac{1}{a}\left(b^2+c^2\right)}+\frac{1}{b+\frac{1}{b}\left(a^2+c^2\right)}+\frac{1}{c+\frac{1}{c}\left(a^2+b^2\right)}\)
\(VT\le\frac{a}{a^2+b^2+c^2}+\frac{b}{a^2+b^2+c^2}+\frac{c}{a^2+b^2+c^2}=\frac{a+b+c}{a^2+b^2+c^2}\)
\(VT\le\frac{a+b+c}{\frac{1}{3}\left(a+b+c\right)^2}=\frac{3}{a+b+c}\le\frac{3}{3\sqrt[3]{abc}}=1\)
Dấu "=" xảy ra khi \(a=b=c=1\)
Ta có: (ÁP DỤNG BĐT CAUCHY SẼ ĐƯỢC):
\(a^4+b^2\ge2\sqrt{a^4b^2}=2a^2b\)
Và: \(b^4+a^2\ge2\sqrt{a^2b^4}=2ab^2\)
=> \(VT\le\frac{a}{2a^2b}+\frac{b}{2ab^2}\)
=> \(VT\le\frac{1}{2ab}+\frac{1}{2ab}\)
=> \(VT\le\frac{2}{2ab}=\frac{1}{ab}\)
=> VẬY TA CÓ ĐPCM.
Dấu "=" xảy ra <=> \(a=b\)
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Ta cm 1 bđt sau:\(a^4+b^4\ge ab\left(a^2+b^2\right)\).Thật vậy:
\(a^4+b^4\ge ab\left(a^2+b^2\right)\Leftrightarrow a^4+b^4-a^3b-ab^3\ge0\)
\(\Leftrightarrow a^3\left(a-b\right)-b^3\left(a-b\right)\ge0\Leftrightarrow\left(a-b\right)^2\left(a^2+ab+b^2\right)\ge0\)Áp dụng: \(T=\frac{a}{b^4+c^4+a}+\frac{c}{a^4+b^4+c}+\frac{b}{c^4+a^4+b}\)
\(T\le\frac{a}{bc\left(b^2+c^2\right)+a}+\frac{c}{ab\left(a^2+b^2\right)+c}+\frac{b}{ac\left(a^2+c^2\right)+b}\)
\(=\frac{a^2}{abc\left(b^2+c^2\right)+a^2}+\frac{c^2}{abc\left(a^2+b^2\right)+c^2}+\frac{b^2}{abc\left(a^2+c^2\right)+b^2}\)
Do abc=1 \(\Rightarrow T\le\frac{a^2+b^2+c^2}{a^2+b^2+c^2}=1."="\Leftrightarrow a=b=c=1\)