Question

[Ngô Thành Chung]

Cho 3 số dương a,b,c. 

CMR : \(\dfrac{a}{\sqrt{a^2+15bc}}+\dfrac{b}{\sqrt{b^2+15ac}}+\dfrac{c}{\sqrt{c^2+15ab}}\ge\dfrac{3}{4}\)

Answer 1

[Trần Tuấn Hoàng]

\(a,b,c>0\)

\(\dfrac{a}{\sqrt{a^2+15bc}}+\dfrac{b}{\sqrt{b^2+15ca}}+\dfrac{c}{\sqrt{c^2+15ab}}\ge\dfrac{3}{4}\)

\(\Leftrightarrow\dfrac{a^2}{a\sqrt{a^2+15bc}}+\dfrac{b^2}{b\sqrt{b^2+15ca}}+\dfrac{c^2}{c\sqrt{c^2+15ab}}\ge\dfrac{3}{4}\)

Áp dụng BĐT Caushy-Schwarz ta được:

\(\dfrac{a^2}{a\sqrt{a^2+15bc}}+\dfrac{b^2}{b\sqrt{b^2+15ca}}+\dfrac{c^2}{c\sqrt{c^2+15ab}}\ge\dfrac{\left(a+b+c\right)^2}{a\sqrt{a^2+15bc}+b\sqrt{b^2+15ca}+c\sqrt{c^2+15ab}}\)

Ta chứng minh rằng:

\(a\sqrt{a^2+15bc}+b\sqrt{b^2+15ca}+c\sqrt{c^2+15ab}\le\dfrac{4}{3}\left(a+b+c\right)^2\)

\(\Leftrightarrow\sqrt{a}\sqrt{a^3+15abc}+\sqrt{b}\sqrt{b^3+15abc}+\sqrt{c}\sqrt{c^3+15abc}\le\dfrac{4}{3}\left(a+b+c\right)^2\)

Áp dụng BĐT Bunhiacopxki ta được:

\(\sqrt{a}\sqrt{a^3+15abc}+\sqrt{b}\sqrt{b^3+15abc}+\sqrt{c}\sqrt{c^3+15abc}\le\sqrt{\left(a+b+c\right)\left(a^3+b^3+c^3+45abc\right)}\)Ta tiếp tục chứng minh:

\(\dfrac{16}{9}\left(a+b+c\right)^3\ge a^3+b^3+c^3+45abc\)

\(\Leftrightarrow\dfrac{16}{9}\left(a^3+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(c+a\right)\right)\ge a^3+b^3+c^3+45abc\)

Áp dụng BĐT AM-GM (Caushy) ta được:

\(\dfrac{16}{9}\left(a^3+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(c+a\right)\right)\ge\dfrac{16}{9}\left(a^3+b^3+c^3+3.2\sqrt{ab}.2.\sqrt{bc}.2.\sqrt{ca}\right)=\dfrac{16}{9}.\left(a^3+b^3+c^3+24abc\right)\)

Ta chứng minh:

\(\dfrac{16}{9}\left(a^3+b^3+c^3+24abc\right)\ge a^3+b^3+c^3+45abc\)

\(\Leftrightarrow\dfrac{16}{9}a^3+\dfrac{16}{9}b^3+\dfrac{16}{9}c^3+\dfrac{16}{9}.24abc\ge a^3+b^3+c^3+45abc\)

\(\Leftrightarrow\dfrac{7}{9}\left(a^3+b^3+c^3\right)\ge\dfrac{7}{3}abc\) (*)

Áp dụng BĐT AM-GM (Caushy) ta được:

\(\dfrac{7}{9}\left(a^3+b^3+c^3\right)\ge\dfrac{7}{9}.3\sqrt[3]{a^3b^3c^3}=\dfrac{7}{3}abc\)

\(\Rightarrow\) (*) đúng.

Vậy BĐT đã được chứng minh. Dấu "=" xảy ra khi \(a=b=c>0\).