Question

Cho a,b,c>0 và a+b+c=2

CMR: \(\sqrt{a^2+\dfrac{1}{a^2}}\)+\(\sqrt{b^2+\dfrac{1}{b^2}}\)+\(\sqrt{c^2+\dfrac{1}{c^2}}\) \(\le\)\(\sqrt{\dfrac{97}{4}}\)

Answer 1

\(P=\sqrt{a^2+\dfrac{1}{a^2}}+\sqrt{b^2+\dfrac{1}{b^2}}+\sqrt{c^2+\dfrac{1}{c^2}}\)

\(\Leftrightarrow\sqrt{\dfrac{97}{4}}P=\sqrt{4+\dfrac{81}{4}}\sqrt{a^2+\dfrac{1}{a^2}}+\sqrt{4+\dfrac{81}{4}}\sqrt{b^2+\dfrac{1}{b^2}}+\sqrt{4+\dfrac{81}{4}}\sqrt{c^2+\dfrac{1}{c^2}}\)

\(\ge\left(2a+\dfrac{9}{2a}\right)+\left(2b+\dfrac{9}{2b}\right)+\left(2c+\dfrac{9}{2c}\right)\)

\(=2\left(a+b+c\right)+\dfrac{9}{2}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)

\(\ge4+\dfrac{9}{2}.\dfrac{9}{a+b+c}=4+\dfrac{81}{4}=\dfrac{97}{4}\)

\(\Rightarrow P\ge\sqrt{\dfrac{97}{4}}\)

PS: Lần sau chép đề cẩn thận nhé bạn.

Answer 2

Toán lớp 10 Chương 4: BẤT ĐẲNG THỨC, BẤT PHƯƠNG TRÌNH §1. Bất đẳng thức

Answer 3

Nếu là \(\ge \) thì easy rồi. Áp dụng BĐT Min....

\(VT=\sqrt{a^2+\dfrac{1}{a^2}}+\sqrt{b^2+\dfrac{1}{b^2}}+\sqrt{c^2+\dfrac{1}{c^2}}\)

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

\(\ge\sqrt{\left(a+b+c\right)^2+\left(\dfrac{9}{a+b+c}\right)^2}\)

\(\ge\sqrt{2^2+\left(\dfrac{9}{2}\right)^2}=\sqrt{\dfrac{97}{4}}=VP\)

Khi \(a=b=c=\frac{2}{3}\)