\(a+b+c=0\)

\(\Leftrightarrow\left(a+b+c\right)^2=0\)

\(\Leftrightarrow a^2+b^2+c^2+2ab+2bc+2ac=0\)

\(\Leftrightarrow a^2+b^2+c^2=-2\left(ab+bc+ac\right)\)

\(\Leftrightarrow\left(a^2+b^2+c^2\right)^2=\left[-2\left(ab+bc+ac\right)\right]^2\)

\(\Leftrightarrow a^4+b^4+c^4+2a^2b^2+2b^2c^2+2a^2c^2=4\left(ab+bc+ac\right)^2\)

\(\Leftrightarrow a^4+b^4+c^4=4\left(ab+bc+ac\right)^2-2a^2b^2-2b^2c^2-2a^2c^2\)

Mà \(\left(ab+bc+ac\right)^2=a^2b^2+b^2c^2+a^2c^2+abc\left(a+b+c\right)\)

\(=a^2b^2+b^2c^2+a^2c^2\)

nên \(a^4+b^4+c^4=4\left(ab+bc+ac\right)^2-2\left(ab+bc+ac\right)^2\)

\(a^4+b^4+c^4=2\left(ab+bc+ac\right)^2\left(đpcm\right)\)

thanks

\(a+b+c=0\)

\(\Leftrightarrow\left(a+b+c\right)^2=0\)

\(\Leftrightarrow a^2+b^2+c^2+2ab+2bc+2ac=0\)

\(\Leftrightarrow a^2+b^2+c^2=-2\left(ab+bc+ac\right)\)

\(\Leftrightarrow\left(a^2+b^2+c^2\right)^2=4\left(ab+bc+ac\right)^2\)

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

\(=4\left(a^2b^2+b^2c^2+a^2c^2+2a^2bc+2ab^2c+2abc^2\right)\)

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

\(=4\left(a^2b^2+b^2c^2+a^2c^2+2abc\left(a+b+c\right)\right)\)

\(\Leftrightarrow a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+a^2c^2\right)\)

\(=4\left(a^2b^2+b^2c^2+a^2c^2\right)\)

\(\Leftrightarrow a^4+b^4+c^4=2\left(a^2b^2+b^2c^2+a^2c^2\right)\)

\(\Leftrightarrow2\left(a^4+b^4+c^4\right)=4\left(a^2b^2+b^2c^2+a^2c^2+2a^2bc+2ab^2c+2abc^2\right)\)

\(\Leftrightarrow2\left(a^4+b^4+c^4\right)=4\left(ab+bc+ac\right)^2\)

\(\Leftrightarrow\left(a^4+b^4+c^4\right)=2\left(ab+bc+ac\right)^2\)

BĐT bị ngược dấu, BĐT đúng phải là:

\(\dfrac{a}{ac+4}+\dfrac{b}{ab+4}+\dfrac{c}{bc+4}\le\dfrac{a^2+b^2+c^2}{16}\)

Bài này thiếu đề. Đề đúng là phải có \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1\) nữa nha bạn.

\(\frac{a^2}{a+bc}+\frac{b^2}{b+ac}+\frac{c^2}{c+ab}\ge\frac{a+b+c}{4}\)

Ta có: \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1\) \(\Rightarrow ab+bc+ac=abc\)

\(VT=\frac{a^2}{a+bc}+\frac{b^2}{b+ac}+\frac{c^2}{c+ab}\)

\(\Rightarrow VT=\frac{a^2.a}{a\left(a+bc\right)}+\frac{b^2.b}{b\left(b+ac\right)}+\frac{c^2.c}{c\left(c+ab\right)}\)

\(\Leftrightarrow VT=\frac{a^3}{a^2+abc}+\frac{b^3}{b^2+abc}+\frac{c^3}{c^2+abc}\)

\(\Leftrightarrow VT=\frac{a^3}{a^2+ab+bc+ac}+\frac{b^3}{b^2+ab+bc+ac}+\frac{c^3}{c^2+ab+bc+ac}\)

\(\Leftrightarrow VT=\frac{a^3}{a\left(a+b\right)+c\left(a+b\right)}+\frac{b^3}{a\left(b+c\right)+b\left(b+c\right)}+\frac{c^3}{c\left(b+c\right)+a\left(b+c\right)}\)

\(\Leftrightarrow VT=\frac{a^3}{\left(a+c\right)\left(a+b\right)}+\frac{b^3}{\left(b+c\right)\left(a+b\right)}+\frac{c^3}{\left(b+c\right)\left(a+c\right)}\)

Áp dụng BĐT Cauchy ta có:

\(\frac{a^3}{\left(a+b\right)\left(a+c\right)}+\frac{a+b}{8}+\frac{a+c}{8}\ge3\sqrt[3]{\frac{a^3}{64}}=\frac{3a}{4}\)

\(\frac{b^3}{\left(a+b\right)\left(b+c\right)}+\frac{a+b}{8}+\frac{b+c}{8}\ge3\sqrt[3]{\frac{b^3}{64}}=\frac{3b}{4}\)

\(\frac{c^3}{\left(b+c\right)\left(a+c\right)}+\frac{b+c}{8}+\frac{a+c}{8}\ge3\sqrt[3]{\frac{c^3}{64}}=\frac{3c}{4}\)

Ta có:

\(\frac{3a}{4}+\frac{3b}{4}+\frac{3c}{4}+\frac{a+b+c}{2}\ge\frac{3}{4}\left(a+b+c\right)\)

\(\Rightarrow\frac{3a}{4}+\frac{3b}{4}+\frac{3c}{4}\ge\frac{3}{4}\left(a+b+c\right)-\frac{1}{2}\left(a+b+c\right)\)

\(\Rightarrow VT\ge\frac{a+b+c}{4}=VP\)

Dấu \("="\) xảy ra \(\Leftrightarrow a=b=c=3\)

\(\RightarrowĐpcm.\)

a. Đề bài sai (thực chất là nó đúng 1 cách hiển nhiên nhưng "dạng" thế này nó sai sai vì ko ai cho kiểu này cả)

Ta có: \(abc=ab+bc+ca\ge3\sqrt[3]{a^2b^2c^2}\Rightarrow abc\ge27\)

\(\Rightarrow a^2+b^2+c^2+5abc\ge a^2+b^2+c^2+5.27>>>>>8\)

b. 

\(4=ab+bc+ca+abc=ab+bc+ca+\sqrt{ab.bc.ca}\le ab+bc+ca+\sqrt{\left(\dfrac{ab+bc+ca}{3}\right)^3}\)

\(\sqrt{\dfrac{ab+bc+ca}{3}}=t\Rightarrow t^3+3t^2-4\ge0\Rightarrow\left(t-1\right)\left(t+2\right)^2\ge0\)

\(\Rightarrow t\ge1\Rightarrow ab+bc+ca\ge3\Rightarrow a+b+c\ge\sqrt{3\left(ab+bc+ca\right)}\ge3\)

- TH1: nếu \(a+b+c\ge4\)

Ta có: \(ab+bc+ca=4-abc\le4\)

\(\Rightarrow P=\left(a+b+c\right)^2-2\left(ab+bc+ca\right)+5abc\ge4^2-2.4+0=8\)

(Dấu "=" xảy ra khi \(\left(a;b;c\right)=\left(2;2;0\right)\) và các hoán vị)

- TH2: nếu \(3\le a+b+c< 4\)

Đặt \(a+b+c=p\ge3;ab+bc+ca=q;abc=r\)

\(P=p^2-2q+5r=p^2-2q+5\left(4-q\right)=p^2-7q+20\)

Áp dụng BĐT Schur:

\(4=q+r\ge q+\dfrac{p\left(4q-p^2\right)}{9}\Leftrightarrow q\le\dfrac{p^3+36}{4p+9}\)

\(\Rightarrow P\ge p^2-\dfrac{7\left(p^3+36\right)}{4p+9}+20=\dfrac{3\left(4-p\right)\left(p-3\right)\left(p+4\right)}{4p+9}+8\ge8\)

(Dấu "=" xảy ra khi \(a=b=c=1\))

Trả lời

Theo đề ra ta có:

\(a+b+c=0\)

\(\Leftrightarrow\left(a+b+c\right)^2=0\)

\(\Leftrightarrow a^2+b^2+c^2+2\left(ab+bc+ac\right)=0\)

\(\Leftrightarrow\left(a^2+b^2+c^2\right)^2=4\left(ab+bc+ca\right)^2\)

\(\Leftrightarrow a^4+b^4+c^4+2\left(a^2\cdot b^2+b^2\cdot c^2+c^2\cdot a^2\right)=4\left(ab+bc+ca\right)^2\)(1)

Lại có:

\(\left(ab+bc+ca\right)^2\)

\(=a^2\cdot b^2+b^2\cdot c^2+c^2\cdot a^2+2bc^2\cdot c+2abc^2+2a^2bc\)

\(=a^2b^2+b^2c^2+c^2a^2=2abc\left(a+b+c\right)\)

\(=a^2b^2+b^2c^2+c^2a^2=2abc\cdot0\)(Do a+b+c=0)

\(=a^2b^2+b^2c^2+c^2a^2\)

Thay \(\left(ab+bc+ca\right)^2=a^2b^2+b^2c^2+c^2a^2\)vào (1); ta có:

\(a^4+b^4+c^4+2\left(ab+bc+ca\right)^2=4\left(ab+bc+ca\right)^2\)

\(\Leftrightarrow a^4+b^4+c^4=2\left(ab+bc+ca\right)^2\)

Vậy \(a,b,c\inℕ\), a+b+c=0 thì \(a^4+b^4+c^4=2\left(ab+bc+ca\right)^2\)(đpcm)

P/s; có gì sai thì góp ý nhé!

Sai đề nha bạn. Không tồn tại 3 số a, b, c > 0 thỏa mãn a + b + c = 0

Đề là cho a,b,c thuôc R nhá 

cho 2 biểu thức mà c/m 1 biểu thức M là sao

Biểu thức N vứt sọt à hay làm cái j v :V

tớ cũng nghĩ vậy nhưng mãi sau mới biết chứng minh M =N rồi chứng minh N >=(a+b+c)/8 để suy ra M  >=(a+b+c)/8