Chủ đề này có 5 trả lời

[conankun]

Cho $a,b,c>0$ và $a+b+c=3$

CMR: $a^4+b^4+c^4\geq a^3+b^3+c^3$

[DOTOANNANG]

$$a^{\,4}\,-\, a^{\,3}\,-\, a\,+\, 1\,+\, b^{\,4}\,-\, b^{\,3}\,-\, b\,+\, 1\,+\, c^{\,4}\,-\, c^{\,3}\,- \,c\,+\, 1\,=$$

 

$$= \,\left ( \,a\,- \,1\, \right )^{\,2}\,\left ( \,a^{\,2}\,+\, a\,+ \,1 \,\right )\,+ \,\left (\,b\,- \,1 \,\right )^{\,2}\,\left (\,b^{\,2}\,+\, b\,+\, 1\,\right )\,+ \,\left (\, c\,- \,1 \,\right )^{\,2}\,\left ( \,c^{\,2}\,+\, c\,+\, 1 \right )\geq  0$$

[PugMath]

Cho $a,b,c>0$ và $a+b+c=3$

CMR: $a^4+b^4+c^4\geq a^3+b^3+c^3

áp dung bđt CBS 

$(\sum{a^4})(\sum{a^2})\geqslant(\sum{a^3})^2$

ta lại có $(a^3+b^3+c^3)(a+b+c)\geqslant (a^2+b^2+c^2)^2$

=>$(\sum{a^4})(\sum{a^2})\geqslant(\sum{a^3})^2\geqslant(\sum{a^3})(\frac{(\sum{a^2})^2}{a+b+c})\geqslant(\sum{a^3})(\sum{a^2})\frac{\frac{(\sum{a})^2}{3}}{\sum{a}}=(\sum{a^3})(\sum{a^2})=>(\sum{a^4})\geqslant(\sum{a^3})$

[conankun]

$$a^{\,4}\,-\, a^{\,3}\,-\, a\,+\, 1\,+\, b^{\,4}\,-\, b^{\,3}\,-\, b\,+\, 1\,+\, c^{\,4}\,-\, c^{\,3}\,- \,c\,+\, 1\,=$$

 

$$= \,\left ( \,a\,- \,1\, \right )^{\,2}\,\left ( \,a^{\,2}\,+\, a\,+ \,1 \,\right )\,+ \,\left (\,b\,- \,1 \,\right )^{\,2}\,\left (\,b^{\,2}\,+\, b\,+\, 1\,\right )\,+ \,\left (\, c\,- \,1 \,\right )^{\,2}\,\left ( \,c^{\,2}\,+\, c\,+\, 1 \right )\geq  0$$

Bạn có cách nào hay hơn ko? Nhớ THCS nha

[conankun]

áp dung bđt CBS 

$(\sum{a^4})(\sum{a^2})\geqslant(\sum{a^3})^2$

ta lại có $(a^3+b^3+c^3)(a+b+c)\geqslant (a^2+b^2+c^2)^2$

=>$(\sum{a^4})(\sum{a^2})\geqslant(\sum{a^3})^2\geqslant(\sum{a^3})(\frac{(\sum{a^2})^2}{a+b+c})\geqslant(\sum{a^3})(\sum{a^2})\frac{\frac{(\sum{a})^2}{3}}{\sum{a}}=(\sum{a^3})(\sum{a^2})=>(\sum{a^4})\geqslant(\sum{a^3})$

Bạn ơi bạn giải giùm mk mà không dùng: $\sum$ nha. Mk cảm ơn.

[Khoa Linh]

Cho $a,b,c>0$ và $a+b+c=3$

CMR: $a^4+b^4+c^4\geq a^3+b^3+c^3$

Áp dụng AM-GM ta có: 

$a^4+a^4+a^4+1\geq 4a^3$

$b^4+b^4+b^4+1\geq 4b^3$

$c^4+c^4+c^4+1\geq 4c^3$

$\Rightarrow 3(a^4+b^4+c^4)\geq 3(a^3+b^3+c^3)+(a^3+b^3+c^3-3)$

mà $(a^3+1+1)+(b^3+1+1)+(c^3+1+1)\geq 3(a+b+c)=9\Leftrightarrow a^3+b^3+c^3\geq 3$

suy ra đpcm