Question

[phạm thảo]

1. Cho a,b,c là các số thực dương thỏa a+b+c=3. Cmr \(\dfrac{a^2}{a+2b^2}+\dfrac{b^2}{b+2c^2}+\dfrac{c^2}{c+2a^2}\ge1\)

2. Cho a,b,c là các số thực dương thỏa \(a^2+b^2+c^2=1\). Cmr: \(\dfrac{a}{1+b^2}+\dfrac{b}{1+c^2}+\dfrac{c}{1+a^2}\ge\dfrac{3}{4}\left(a\sqrt{a}+b\sqrt{b}+c\sqrt{c}\right)^2\)

3.Cho a,b,c là các số thực dương thỏa \(a^2+b^2+c^2=3\). Cmr:\(\sqrt{\dfrac{a^2}{b+b^2+c}}+\sqrt{\dfrac{b^2}{c+c^2+a}}+\sqrt{\dfrac{c^2}{a+a^2+b}}\le3\)

Answer 1

[Akai Haruma]

Bài 1:

Ta có:

\(\text{VT}=\frac{a^2}{a+2b^2}+\frac{b^2}{b+2c^2}+\frac{c^2}{c+2a^2}\)

\(=a-\frac{2ab^2}{a+2b^2}+b-\frac{2bc^2}{b+2c^2}+c-\frac{2ca^2}{c+2a^2}=(a+b+c)-2\left(\frac{ab^2}{a+2b^2}+\frac{bc^2}{b+2c^2}+\frac{ca^2}{c+2a^2}\right)\)

\(=3-2M(*)\)

Áp dụng BĐT Cauchy ta có:

\(M=\frac{ab^2}{a+b^2+b^2}+\frac{bc^2}{b+c^2+c^2}+\frac{ca^2}{c+a^2+a^2}\leq \frac{ab^2}{3\sqrt[3]{ab^4}}+\frac{bc^2}{3\sqrt[3]{bc^4}}+\frac{ca^2}{3\sqrt[3]{ca^4}}\)

\(\Leftrightarrow M\leq \frac{1}{3}(\sqrt[3]{a^2b^2}+\sqrt[3]{b^2c^2}+\sqrt[3]{c^2a^2})\)

Tiếp tục áp dụng BĐT Cauchy:

\(\sqrt[3]{a^2b^2}+\sqrt[3]{b^2c^2}+\sqrt[3]{c^2a^2}\leq \frac{ab+ab+1}{3}+\frac{bc+bc+1}{3}+\frac{ca+ca+1}{3}=\frac{2(ab+bc+ac)+3}{3}\)

Mà \(ab+bc+ac\leq \frac{(a+b+c)^2}{3}=3\) (quen thuộc)

\(\Rightarrow M\leq \frac{1}{3}.\frac{2.3+3}{3}=1(**)\)

Từ \((*);(**)\Rightarrow \text{VT}\geq 3-2.1=1\)

(đpcm)

Dấu bằng xảy ra khi $a=b=c=1$

Answer 2

[Akai Haruma]

Bài 2:

Áp dụng BĐT Cauchy -Schwarz:

\(\text{VT}=\frac{a^3}{a^2+a^2b^2}+\frac{b^3}{b^2+b^2c^2}+\frac{c^3}{c^2+a^2c^2}\geq \frac{(a\sqrt{a}+b\sqrt{b}+c\sqrt{c})^2}{a^2+a^2b^2+b^2+b^2c^2+c^2+c^2a^2}\)

hay:

\(\text{VT}\geq \frac{(a\sqrt{a}+b\sqrt{b}+c\sqrt{c})^2}{1+a^2b^2+b^2c^2+c^2a^2}(*)\)

Mặt khác, theo BĐT Cauchy ta dễ thấy:

\(a^4+b^4+c^4\geq a^2b^2+b^2c^2+c^2a^2\)

\(\Rightarrow (a^2+b^2+c^2)^2\geq 3(a^2b^2+b^2c^2+c^2a^2)\)

\(\Leftrightarrow 1\geq 3(a^2b^2+b^2c^2+c^2a^2)\Rightarrow a^2b^2+b^2c^2+c^2a^2\leq \frac{1}{3}(**)\)

Từ \((*);(**)\Rightarrow \text{VT}\geq \frac{(a\sqrt{a}+b\sqrt{b}+c\sqrt{c})^2}{1+\frac{1}{3}}=\frac{3}{4}(a\sqrt{a}+b\sqrt{b}+c\sqrt{c})^2\)

Ta có đpcm

Dấu bằng xảy ra khi \(a=b=c=\frac{1}{\sqrt{3}}\)