Vì \(:a>b\) nên \(2a>2b\)

\(\rightarrow2a-1>2b-1\)

\(\rightarrow2a-1+1>2b-1+1\)

\(\rightarrow2a-1+1>2b-1+1-1\)

\(\rightarrow2a>2b-1\)

\(\rightarrowĐPCM\)

a > b

=> 2a - 1 > 2b - 1

=> 2a > 2b - 1

1.

\(2a^2b^2+2b^2c^2+2c^2a^2-a^4-b^4-c^4>0\\ \Leftrightarrow a^4+b^4+c^4-2a^2b^2-2b^2c^2-2c^2a^2< 0\\ \Leftrightarrow\left(a^4+b^4+c^4+2a^2b^2-2b^2c^2-2c^2a^2\right)-4a^2b^2< 0\\ \Leftrightarrow\left(a^2+b^2-c^2\right)^2-4a^2b^2< 0\\ \Leftrightarrow\left(a^2+b^2-c^2-2ab\right)\left(a^2+b^2-c^2+2ab\right)< 0\\ \Leftrightarrow\left[\left(a-b\right)^2-c^2\right]\left[\left(a+b\right)^2-c^2\right]< 0\\ \Leftrightarrow\left(a-b+c\right)\left(a-b-c\right)\left(a+b-c\right)\left(a+b+c\right)< 0\left(1\right)\)

Vì a,b,c là độ dài 3 cạnh của 1 tg nên \(\left\{{}\begin{matrix}a+c>b\\a-b< c\\a+b>c\\a+b+c>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a-b+c>0\\a-b-c< 0\\a+b-c>0\\a+b+c>0\end{matrix}\right.\)

Do đó \(\left(1\right)\) luôn đúng (do 3 dương nhân 1 âm ra âm)

Từ đó ta được đpcm

2.

\(a,Sửa:a^6+a^4+a^2b^2+b^4-b^6\\ =\left(a^6-b^6\right)+\left(a^4+b^4+a^2b^2\right)\\ =\left(a^2-b^2\right)\left(a^4+a^2b^2+b^4\right)+\left(a^4+b^4+a^2b^2\right)\\ =\left(a^2-b^2+1\right)\left(a^4+a^2b^2+b^4\right)\\ =\left[\left(a^2+b^2\right)^2-a^2b^2\right]\left(a^2-b^2+1\right)\\ =\left(a^2-ab+b^2\right)\left(a^2+ab+b^2\right)\left(a^2-b^2+1\right)\\ b,=\left(a^3+b^3\right)-1+3ab\\ =\left(a+b\right)^3-3ab\left(a+b\right)-1+3ab\\ =\left(a+b-1\right)\left(a^2+2ab+b^2+a+b+1\right)-3ab\left(a+b-1\right)\\ =\left(a+b-1\right)\left(a^2+b^2+1+a+b-ab\right)\)

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

a + b = c => (a + b)² = c² <=> a²+ b² + 2ab = c² 
=> c^4 = (a² + b² + 2ab)² 
=> c^4 = a^4 + b^4 + 6a²b² + 4a^3.b + 4a.b^3 

vậy: a^4 + b^4 + c^4 = 2a^4 + 2b^4 + 6a²b² + 4a^3.b + 4a.b^3 
= 2a^4 + 2a²b² + 4a^3.b + 2b^4 + 2a²b² + 4a.b^3 + 2a²b² 
= 2a²(a² + b² + 2ab) + 2b²(b² + a² + 2ab) + 2a²b² 
= 2a²(a + b)² + 2b²(a + b)² + 2a²b² 
= 2a²b² + 2(a + b)²(a² + b²) 
= 2a²b² + 2c²(a² +b²) 
= 2a²b² + 2b²c² + 2c²a² (đpcm) 

 gt: a + b = c => (a + b)² = c² <=> a²+ b² + 2ab = c² 
=> c^4 = (a² + b² + 2ab)² 
=> c^4 = a^4 + b^4 + 6a²b² + 4a^3.b + 4a.b^3 

vậy: a^4 + b^4 + c^4 = 2a^4 + 2b^4 + 6a²b² + 4a^3.b + 4a.b^3 
= 2a^4 + 2a²b² + 4a^3.b + 2b^4 + 2a²b² + 4a.b^3 + 2a²b² 
= 2a²(a² + b² + 2ab) + 2b²(b² + a² + 2ab) + 2a²b² 
= 2a²(a + b)² + 2b²(a + b)² + 2a²b² 
= 2a²b² + 2(a + b)²(a² + b²) 
= 2a²b² + 2c²(a² +b²) 
= 2a²b² + 2b²c² + 2c²a² (đpcm) 

a: 3a+2b>=3b+2a

=>3a-2a>=3b-2b

=>a>=b(đúng)

b: =>a^2-2ab+b^2<=2a^2+2b^2

=>2a^2+2b^2-a^2+2ab-b^2>=0

=>(a+b)^2>=0(luôn đúng)

c: =>5a^2+5b^2>=4a^2-4ab+b^2

=>a^2+4ab+4b^2>=0

=>(a+2b)^2>=0(luôn đúng)

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