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Lời giải:
\(\frac{a^2+bc}{b+c}+\frac{b^2+ac}{c+a}+\frac{c^2+ab}{a+b}\geq a+b+c\)
\(\Leftrightarrow \frac{a^2+bc}{b+c}-c+\frac{b^2+ac}{a+c}-a+\frac{c^2+ab}{a+b}-b\geq 0\)
\(\Leftrightarrow \frac{a^2-c^2}{b+c}+\frac{b^2-a^2}{a+c}+\frac{c^2-b^2}{a+b}\geq 0\)
\(\Leftrightarrow a^2\left(\frac{1}{b+c}-\frac{1}{a+c}\right)+b^2\left(\frac{1}{a+c}-\frac{1}{a+b}\right)+c^2\left(\frac{1}{a+b}-\frac{1}{b+c}\right)\geq 0\)
\(\Leftrightarrow \frac{a^2(a-b)(a+b)+b^2(b-c)(b+c)+c^2(c-a)(c+a)}{(a+b)(b+c)(c+a)}\geq 0\)
\(\Leftrightarrow a^2(a^2-b^2)+b^2(b^2-c^2)+c^2(c^2-a^2)\geq 0\)
\(\Leftrightarrow a^4+b^4+c^4-(a^2b^2+b^2c^2+c^2a^2)\geq 0\)
\(\Leftrightarrow \frac{(a^2-b^2)^2+(b^2-c^2)^2+(c^2-a^2)^2}{2}\geq 0\) (luôn đúng)
Do đó ta có đpcm
Dấu bằng xảy ra khi $a=b=c$
\(M=\frac{a^3+b^3}{a^2+ab+b^2}+\frac{b^3+c^3}{b^2+bc+c^2}+\frac{c^3+a^3}{c^2+ac+a^2}\)
\(=\left(\frac{a^3+b^3}{a^2+ab+b^2}-b+a\right)+\left(\frac{b^3+c^3}{b^2+bc+c^2}-c+b\right)+\left(\frac{c^3+a^3}{c^2+ac+a^2}-a+c\right)\)
\(=2\left(\frac{a^3}{a^2+ab+b^2}+\frac{b^3}{b^2+bc+c^2}+\frac{c^3}{c^2+ac+c^2}\right)\)
\(=2....\) ( đề thiếu )
Áp dụng BĐT Cauchy-Schwarz dạng Engel ta có:
\(VT=\dfrac{a^6}{a^3+a^2b+ab^2}+\dfrac{b^6}{b^3+b^2c+bc^2}+\dfrac{c^6}{c^3+ac^2+a^2c}\)
\(\ge\dfrac{\left(a^3+b^3+c^3\right)^2}{a^3+a^2b+ab^2+b^3+b^2c+bc^2+c^3+ac^2+a^2c}\)
\(=\dfrac{\left(a^3+b^3+c^3\right)^2}{\left(a+b+c\right)\left(a^2+b^2+c^2\right)}\). Cần chứng minh BĐT
\(\dfrac{\left(a^3+b^3+c^3\right)^2}{\left(a+b+c\right)\left(a^2+b^2+c^2\right)}\ge\dfrac{a^3+b^3+c^3}{3}\)
\(\Leftrightarrow\dfrac{a^3+b^3+c^3}{\left(a+b+c\right)\left(a^2+b^2+c^2\right)}\ge\dfrac{1}{3}\Leftrightarrow3\left(a^3+b^3+c^3\right)\ge\left(a+b+c\right)\left(a^2+b^2+c^2\right)\)
Lại xài BĐT Holder ta có:
\(\left(1+1+1\right)\left(1+1+1\right)\left(a^3+b^3+c^3\right)\ge\left(a+b+c\right)^3\)
\(\Rightarrow3\left(a^3+b^3+c^3\right)\ge\dfrac{\left(a+b+c\right)^3}{3}\ge\left(a+b+c\right)\left(a^2+b^2+c^2\right)\)
\(\Leftrightarrow\dfrac{\left(a+b+c\right)^2}{3}\ge a^2+b^2+c^2\)
BĐT cuối nên ta có cả bài này sai. Ai có cách khác hay soi lỗi hộ thì tks trước :v
Câu 1:
Ta có: \(\left(\dfrac{a+b}{2}\right)^2\ge ab\)
\(\Leftrightarrow\dfrac{\left(a+b\right)^2}{2^2}-ab\ge0\)
\(\Leftrightarrow\dfrac{a^2+2ab+b^2-4ab}{4}\ge0\)
\(\Leftrightarrow\dfrac{a^2-2ab+b^2}{4}\ge0\)
\(\Leftrightarrow\dfrac{\left(a-b\right)^2}{4}\ge0\)
Vì \(\left(a-b\right)^2\ge0\forall a,b\)
\(\Rightarrow\dfrac{\left(a-b\right)^2}{4}\ge0\forall a,b\)
\(\Rightarrow\left(\dfrac{a+b}{2}\right)^2\ge ab\) (1)
Ta có: \(\dfrac{a^2+b^2}{2}\ge\left(\dfrac{a+b}{2}\right)^2\)
\(\Leftrightarrow\dfrac{a^2+b^2}{2}-\dfrac{\left(a+b\right)^2}{4}\ge0\)
\(\Leftrightarrow\dfrac{2a^2-2b^2-a^2-2ab-b^2}{4}\ge0\)
\(\Leftrightarrow\dfrac{a^2-2ab-b^2}{4}\ge0\)
\(\Leftrightarrow\dfrac{\left(a-b\right)^2}{4}\ge0\)
Vì \(\left(a-b\right)^2\ge0\forall a,b\)
\(\Rightarrow\dfrac{\left(a-b\right)^2}{4}\ge0\forall a,b\)
\(\Rightarrow\dfrac{a^2+b^2}{2}\ge\left(\dfrac{a+b}{2}\right)^2\) (2)
Từ (1) và (2) \(\Rightarrow ab\le\left(\dfrac{a+b}{2}\right)^2\le\dfrac{a^2+b^2}{2}\)
5 , a3+b3+c3\(\ge\) 3abc
\(\Leftrightarrow\) a3+3a2b+3ab2+b3+c3-3a2b-3ab2-3abc\(\ge\) 0
\(\Leftrightarrow\) (a+b)3+c3-3ab(a+b+c) \(\ge0\)
\(\Leftrightarrow\) (a+b+c)(a2+2ab+b2-ac-bc+c2)-3ab(a+b+c) \(\ge0\)
\(\Leftrightarrow\) (a+b+c)(a2+b2+c2-ab-bc-ca)\(\ge0\) (1)
ta co : a,b,c>0 \(\Rightarrow\)a+b+c>0 (2)
(a-b)2+(b-c)2+(c-a)2\(\ge0\)
<=> 2a2+2b2+2c2-2ac-2cb-2ab\(\ge0\)
<=>a2+b2+c2-ab-bc-ac\(\ge\) 0 (3)
Từ (1)(2)(3)=> pt luôn đúng
Đặt vế trái của BĐT cần chứng minh là P
Ta có:
\(\dfrac{a^2}{a^2+b^2+c^2-bc}=\dfrac{2a^2}{2a^2+b^2+c^2+\left(b-c\right)^2}\le\dfrac{2a^2}{2a^2+b^2+c^2}=\dfrac{2a^2}{a^2+b^2+a^2+c^2}\)
\(\le\dfrac{1}{2}\left(\dfrac{a^2}{a^2+b^2}+\dfrac{a^2}{a^2+c^2}\right)\)
Tương tự:
\(\dfrac{b^2}{a^2+b^2+c^2-ac}\le\dfrac{1}{2}\left(\dfrac{b^2}{a^2+b^2}+\dfrac{b^2}{b^2+c^2}\right)\)
\(\dfrac{c^2}{a^2+b^2+c^2-ab}\le\dfrac{1}{2}\left(\dfrac{c^2}{a^2+c^2}+\dfrac{c^2}{b^2+c^2}\right)\)
Cộng vế với vế:
\(P\le\dfrac{1}{2}\left(\dfrac{a^2+b^2}{a^2+b^2}+\dfrac{b^2+c^2}{b^2+c^2}+\dfrac{c^2+a^2}{a^2+c^2}\right)=\dfrac{3}{2}\)
Dấu "=" xảy ra khi \(a=b=c\)
Bài 1:
Từ \(a+b+c=0\) ta có:
\(B=\frac{a^2}{a^2-b^2-c^2}+\frac{b^2}{b^2-c^2-a^2}+\frac{c^2}{c^2-b^2-a^2}\)
\(=\frac{a^2}{(-b-c)^2-b^2-c^2}+\frac{b^2}{(-c-a)^2-c^2-a^2}+\frac{c^2}{(-b-a)^2-b^2-a^2}\)
\(=\frac{a^2}{2bc}+\frac{b^2}{2ac}+\frac{c^2}{2ab}=\frac{a^3+b^3+c^3}{2abc}\)
Lại có:
\(a^3+b^3+c^3=(a+b)^3-3ab(a+b)+c^3=(-c)^3-3ab(-c)+c^3\)
\(=-c^3+3abc+c^3=3abc\)
Do đó \(B=\frac{3abc}{2abc}=\frac{3}{2}\)
Bài 2:
Lấy P-Q ta có:
\(P-Q=\left(\frac{a^3}{a^2+ab+b^2}+\frac{b^3}{b^2+bc+c^2}+\frac{c^3}{c^2+ca+a^2}\right)-\left(\frac{b^3}{a^2+ab+b^2}+\frac{c^3}{b^2+bc+c^2}+\frac{a^3}{c^2+ca+a^2}\right)\)
\(P-Q=\frac{a^3-b^3}{a^2+ab+b^2}+\frac{b^3-c^3}{b^2+bc+c^2}+\frac{c^3-a^3}{c^2+ac+a^2}\)
\(P-Q=\frac{(a-b)(a^2+ab+b^2)}{a^2+ab+b^2}+\frac{(b-c)(b^2+bc+c^2)}{b^2+bc+c^2}+\frac{(c-a)(c^2+ac+a^2)}{c^2+ac+a^2}\)
\(P-Q=(a-b)+(b-c)+(c-a)=0\Rightarrow P=Q\)
Ta có đpcm.
e)
\(\dfrac{a^2+b^2+c^2}{3}\ge\left(\dfrac{a+b+c}{3}\right)^2\)
\(\Leftrightarrow3\left(a^2+b^2+c^2\right)\ge a^2+b^2+c^2+2\left(ab+bc+ca\right)\)
\(\Leftrightarrow2\left(a^2+b^2+c^2\right)\ge2\left(ab+bc+ac\right)\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2ac-2bc\ge0\)
\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(a^2-2ac+c^2\right)+\left(b^2-2bc+c^2\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(a-c\right)^2+\left(b-c\right)^2\ge0\) ( luôn đúng)
=> ĐPCM
Lời giải:
Áp dụng BĐT Cauchy_ Schwarz ta có:
\(\text{VT}=\frac{a^6}{a^3+a^2b+ab^2}+\frac{b^6}{b^3+b^2c+bc^2}+\frac{c^6}{c^3+c^2a+ca^2}\)
\(\geq \frac{(a^3+b^3+c^3)^2}{a^3+a^2b+ab^2+b^3+b^2c+bc^2+c^3+c^2a+ca^2}\)
\(\Leftrightarrow \text{VT}\geq \frac{(a^3+b^3+c^3)^2}{a^3+b^3+c^3+ab(a+b)+bc(b+c)+ac(a+c)}\) (I)
Áp dụng BĐT Am-Gm ta có:
\(\left\{\begin{matrix} a^3+a^3+b^3\geq 3a^2b\\ b^3+b^3+c^3\geq 3b^2c\\ c^3+c^3+a^3\geq 3c^2a\end{matrix}\right.\Rightarrow 3(a^3+b^3+c^3)\geq 3(a^2b+b^2c+c^2a)\)
\(\Leftrightarrow a^3+b^3+c^3\geq a^2b+b^2c+c^2a\) (1)
Tương tự:
\(\left\{\begin{matrix} a^3+b^3+b^3\geq 3ab^2\\ b^3+c^3+c^3\geq 3bc^2\\ c^3+a^3+a^3\geq 3ca^2\end{matrix}\right.\Rightarrow 3(a^3+b^3+c^3)\geq 3(ab^2+bc^2+ca^2)\)
\(\Leftrightarrow a^3+b^3+c^3\geq ab^2+bc^2+ca^2(2)\)
Từ \((1);(2)\Rightarrow 2(a^3+b^3+c^3)\geq ab(a+b)+bc(b+c)+ac(c+a)\)
\(\Rightarrow a^3+b^3+c^3+ab(a+b)+bc(b+c)+ac(c+a)\leq 3(a^3+b^3+c^3)\) (II)
Từ \((I);(II)\Rightarrow \text{VT}\geq \frac{(a^3+b^3+c^3)^2}{a^3+b^3+c^3+ab(a+b)+bc(b+c)+ac(a+c)}\geq \frac{(a^3+b^3+c^3)^2}{3(a^3+b^3+c^3)}\)
\(\Leftrightarrow \text{VT}\geq \frac{a^3+b^3+c^3}{3}\) (đpcm)
Dấu bằng xảy ra khi \(a=b=c\)