We have:\(\hept{\begin{cases}a^2+b^2+c^2=\frac{1}{3}\\a,b,c>0\end{cases}\Rightarrow0< a,b,c< \frac{1}{\sqrt{3}}}\)

We prove to:

\(4x+\frac{2}{3x}\ge-3x^2+\frac{11}{3}\)  with  \(0< x< \frac{1}{\sqrt{3}}\)

\(\Leftrightarrow4x+\frac{2}{3x}+3x^2-\frac{11}{3}\ge0\)

\(\Leftrightarrow9x^3+12x^2-11x+2\ge0\)

\(\Leftrightarrow\left(3x+1\right)^2\left(x+2\right)\ge0\)   Always true to all \(0< x< \frac{1}{\sqrt{3}}\) 

\(\Rightarrow VT\ge-3a^2+\frac{11}{3}-3b^2+\frac{11}{3}-3c^2+\frac{11}{3}\)

\(=-3\left(a^2+b^2+c^2\right)+11=-3.\frac{1}{3}+11=10\) \(\left(đpcm\right)\)

Đặt biểu thức trên là \(A\)

Ta có : \(A=\left(4a+\frac{2}{3a}\right)+\left(4b+\frac{2}{3b}\right)+\left(4c+\frac{2}{3c}\right)\)

Cần chứng minh \(4a+\frac{2}{3a}\ge-3a^2+\frac{11}{3}\) (*)

Thật vậy \(BĐT\Leftrightarrow4a+\frac{2}{3a}+3a^2-\frac{11}{3}\ge0\)

\(\Leftrightarrow\frac{12a^2+2+9a^3-11a}{3a}\ge0\Leftrightarrow\frac{\left(a+2\right)\left(3a-1\right)^2}{3a}\ge0\) (luôn đúng)

Tương tự : \(4b+\frac{2}{3b}\ge-3b^2+\frac{11}{3}\)   và \(4c+\frac{2}{3c}\ge-3c^2+\frac{11}{3}\)

Cộng các bất dẳng thức vừa CM đc ta có :

\(A\ge-3\left(a^2+b^2+c^2\right)+\frac{11}{3}.3=-3.\frac{1}{3}+11=10\)

Dấu "=" xảy ra \(\Leftrightarrow a=b=c=\frac{1}{3}\)

Đặt \(x=\frac{2a}{b+c};y=\frac{2b}{c+a};z=\frac{2c}{a+b}\) Thì bài toán thành chứng minh

\(3\left(\sqrt{\frac{a+b}{2c}}+\sqrt{\frac{b+c}{2a}}+\sqrt{\frac{c+a}{2b}}\right)^2\ge\frac{8\left(a+b+c\right)^3}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)

Áp dụng holder ta có:

\(\left(\sqrt{\frac{a+b}{2c}}+\sqrt{\frac{b+c}{2a}}+\sqrt{\frac{c+a}{2b}}\right)^2\left(2c\left(a+b\right)^2+2a\left(b+c\right)^2+2b\left(c+a\right)^2\right)\)

\(\ge\left[\left(a+b\right)+\left(b+c\right)+\left(c+a\right)\right]^3=8\left(a+b+c\right)^3\)

\(\Rightarrow VT\ge3.\frac{8\left(a+b+c\right)^3}{2a\left(b+c\right)^2+2b\left(c+a\right)^2+2c\left(a+b\right)^2}\)

Từ đây ta cần chứng minh:

\(3.\frac{8\left(a+b+c\right)^3}{2a\left(b+c\right)^2+2b\left(c+a\right)^2+2c\left(a+b\right)^2}\ge\frac{8\left(a+b+c\right)^3}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)

\(\Leftrightarrow2a\left(b+c\right)^2+2b\left(c+a\right)^2+2c\left(a+b\right)^2\le3\left(a+b\right)\left(b+c\right)\left(c+a\right)\)

\(\Leftrightarrow a\left(b-c\right)^2+b\left(c-a\right)^2+c\left(a-b\right)^2\ge0\)( đúng )

Vậy có ĐPCM

đi ,nt ,mình giải cho

nt là gì

Lời giải:

Áp dụng BĐT Cauchy-Schwarz ta có:

\(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{2\sqrt[3]{abc}}=\frac{c^2}{c^2(a+b)}+\frac{a^2}{a^2(b+c)}+\frac{b^2}{b^2(c+a)}+\frac{(\sqrt[3]{abc})^2}{2abc}\)

\(\geq \frac{(c+a+b+\sqrt[3]{abc})^2}{c^2(a+b)+a^2(b+c)+b^2(c+a)+2abc}=\frac{(a+b+c+\sqrt[3]{abc})^2}{(a+b)(b+c)(c+a)}\)

Ta có đpcm

Dấu "=" xảy ra khi $a=b=c$

\(VT=\frac{\left(\sqrt[3]{abc}\right)^2}{2abc}+\Sigma\frac{a^2}{a^2\left(b+c\right)}\ge\frac{\left(a+b+c+\sqrt[3]{abc}\right)^2}{\Sigma a^2\left(b+c\right)+2abc}=\frac{\left(a+b+c+\sqrt[3]{abc}\right)^2}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)

Áp dụng BĐT AM-GM ta có \(\frac{1^2}{a\left(a+b\right)}+\frac{1^2}{b\left(b+c\right)}+\frac{1^2}{c\left(c+a\right)}\ge\)

\(\ge\frac{\left(1+1+1\right)^2}{a\left(a+b\right)+b\left(b+c\right)+c\left(c+a\right)}=\frac{9}{a\left(a+b\right)+b\left(b+c\right)+c\left(c+a\right)}\ge\)

\(\ge\frac{9}{3.\sqrt[3]{abc\left(a+b\right)\left(b+c\right)\left(c+a\right)}}\)

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