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Lời giải:
Bạn nhớ tới bổ đề sau: Với $a,b>0$ thì $a^3+b^3\geq ab(a+b)$.
Áp dụng vào bài:
$5a^3-b^3\leq 5a^3-[ab(a+b)-a^3]=6a^3-ab(a+b)$
$\Rightarrow \frac{5a^3-b^3}{ab+3a^2}\leq \frac{6a^3-ab(a+b)}{ab+3a^2}=\frac{6a^2-ab-b^2}{3a+b}=\frac{(3a+b)(2a-b)}{3a+b}=2a-b$
Tương tự:
$\frac{5b^3-c^3}{bc+3b^2}\leq 2b-c; \frac{5c^3-a^3}{ca+3c^2}\leq 2c-a$
Cộng theo vế:
$\Rightarrow \text{VT}\leq a+b+c=3$
Ta có đpcm
Dấu "=" xảy ra khi $a=b=c=1$
Đề bị lỗi hiển thị hay sao ấy, mình không nhìn thấy BĐT/ đẳng thức bạn muốn chứng minh.
\(VT=\sum\dfrac{a^2}{5a^2+b^2+c^2+2bc}=\sum\dfrac{a^2}{\left(2a^2+bc\right)+\left(2a^2+bc\right)+a^2+b^2+c^2}\)
\(\le\sum\dfrac{a^2}{9}\left(\dfrac{2}{2a^2+bc}+\dfrac{1}{a^2+b^2+c^2}\right)=\dfrac{1}{9}+\sum\dfrac{2a^2}{9\left(2a^2+bc\right)}\)
\(=\dfrac{4}{9}-\dfrac{1}{9}\left(\dfrac{bc}{2a^2+bc}+\dfrac{ac}{2b^2+ac}+\dfrac{ab}{2c^2+ab}\right)\)
\(\le\dfrac{4}{9}-\dfrac{1}{9}.\dfrac{\left(ab+bc+ca\right)^2}{\left(ab+bc+ca\right)^2}=\dfrac{1}{3}\)
Dấu = xảy ra khi a=b=c
\(\dfrac{\left(b+c\right)^2}{5a^2+\left(b+c\right)^2}+\dfrac{\left(c+a\right)^2}{5b^2+\left(c+a\right)^2}+\dfrac{\left(a+b\right)^2}{5c^2+\left(a+b\right)}\ge\dfrac{4}{3}\)
\(\Leftrightarrow\dfrac{-20a^2+10bc+5b^2+c^2}{9\left(5a^2+\left(b+c\right)^2\right)}+\dfrac{-20b^2+10ac+5c^2+5a^2}{9\left(5b^2+\left(c+a\right)^2\right)}+\dfrac{-20c^2+10ab+5a^2+5b^2}{9\left(5c^2+\left(a+b\right)\right)}\ge0\)
\(\Leftrightarrow\sum_{cyc}\dfrac{\left(c-a\right)\left(10a+5b+5c\right)-\left(a-b\right)\left(10a+5b+5c\right)}{9\left(5a^2+\left(b+c\right)^2\right)}\ge0\)
\(\Leftrightarrow\sum_{cyc}\left(\dfrac{-\left(a-b\right)\left(10a+5b+5c\right)}{9\left(5a^2+\left(b+c\right)^2\right)}+\dfrac{\left(a-b\right)\left(10b+5a+5c\right)}{9\left(5b^2+\left(a+c\right)^2\right)}\right)\ge0\)
\(\Leftrightarrow\sum_{cyc}\left(\left(a-b\right)\left(\dfrac{10b+5a+5c}{9\left(5b^2+\left(a+c\right)^2\right)}-\dfrac{10a+5b+5c}{9\left(5a^2+\left(b+c\right)^2\right)}\right)\right)\ge0\)
\(\Leftrightarrow\sum_{cyc}\left(\left(a-b\right)^2\dfrac{5\left(a^2+b^2-c^2+4ab\right)}{3\left(a^2+2ac+5b^2+c^2\right)\left(5a^2+b^2+2bc+c^2\right)}\right)\ge0\)
Dau "=" khi \(a=b=c\)
Ta chứng minh bổ đề sau:
\(\dfrac{5b^3-a^3}{ab+3b^2}\le2b-a\)
\(\Leftrightarrow5b^3-a^3\le\left(2b-a\right)\left(ab+3b^2\right)\)
\(\Leftrightarrow5b^3-a^3\le2ab^2+6b^3-a^2b-3b^2a\)
\(\Leftrightarrow a^3+b^3-a^2b-b^2a\ge0\)
\(\Leftrightarrow\left(a+b\right)\left(a^2-ab+b^2\right)-ab\left(a+b\right)\ge0\)
\(\Leftrightarrow\left(a+b\right)\left(a^2-2ab+b^2\right)\ge0\)
\(\Leftrightarrow\left(a+b\right)\left(a-b\right)^2\ge0\)
Bất đẳng thức cuối luôn đúng, vậy ta có
\(M\le2a-b+2b-c+2c-a=a+b+c\)Chứng minh hoàn tất. Đẳng thức xảy ra khi \(a=b=c\)
Ta có \(a+b^2\le\dfrac{a^2+1}{2}+b^2=\dfrac{a^2+2b^2+1}{2}\)
\(\Rightarrow\dfrac{2a^2}{a+b^2}\ge\dfrac{4a^2}{a^2+2b^2+1}=\dfrac{4a^4}{a^4+2b^2a^2+a^2}\). Lập 2 BĐT tương tự rồi áp dụng bất đẳng thức BCS, ta có:
\(\dfrac{2a^2}{a+b^2}+\dfrac{2b^2}{b+c^2}+\dfrac{2c^2}{c+a^2}\ge\dfrac{\left(2a^2+2b^2+2c^2\right)^2}{a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+c^2a^2\right)+a^2+b^2+c^2}\) \(=\dfrac{4\left(a^2+b^2+c^2\right)^2}{\left(a^2+b^2+c^2\right)^2+3}\)\(=\dfrac{4.3^2}{3^2+3}=3\).
Mà \(a+b+c\le\sqrt{3\left(a^2+b^2+c^2\right)}=3\) nên ta có đpcm. ĐTXR \(\Leftrightarrow a=b=c=1\)
Đề bài sai
Đề đúng: \(\dfrac{1}{\sqrt{a}+2\sqrt{b}+3}+\dfrac{1}{\sqrt{b}+2\sqrt{c}+3}+\dfrac{1}{\sqrt{c}+2\sqrt{a}+3}\le\dfrac{1}{2}\)
Chuẩn hóa: a+b+c=3k
\(\Rightarrow\)\(\dfrac{a}{k}+\dfrac{b}{k}+\dfrac{c}{k}=3\)
Đặt (\(\dfrac{a}{k};\dfrac{b}{k};\dfrac{c}{k}\))\(\Rightarrow\left(x;y;z\right)\);x+y+z=3
ĐPCM\(\Leftrightarrow\)\(\sum\dfrac{19y^3-x^3}{xy+5y^2}\le3\left(x+y+z\right)\)
Ta CM BĐT:
\(\dfrac{19y^3-x^3}{xy+5y^2}\le4y-x\Leftrightarrow-\dfrac{\left(y-x\right)^2\left(x+y\right)}{xy+5y^2}\le0\)(đúng)
CMTT\(\Rightarrow\)ĐPCM